Problem about functional equation (Bulgarian selection team test)












8














This problem was taken from the bulgarian selection team test for the 47th IMO and appeared in a chinese magazine, I came across it in my own training.



http://www.math.ust.hk/excalibur/v10_n4.pdf



The problem is as follows:



Consider $f:Arightarrow A$ where $A$ is the set of non-zero real numbers. Find all such functions that satisfy the following functional equation for all $x$ and $y$ in $A$:



$$f(x^{2}+y) = f(x)^{2} + frac{f(xy)}{f(x)}$$



I suspect the only function is the identity, but the problem has proved itself to be too difficult for me, I've pondered about it for various hours. Any hint or solution is welcome! Thanks.










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  • wait...so you are suggesting $f(x) = a^2 + 1$, i.e. f(x) is a constant function? that doesn't work... $f(x^2+y)=a^2+1$ but $f(x)^2 + frac{f(xy)}{f(x)} = a^4 + 2a^2 + 2$
    – user430825
    Jun 13 '14 at 1:20












  • ah I see! thanks for clarification. Just one problem I see, the original formulation $alpha = alpha^{2} + 1$ only has complex solutions, which are outside of the codomain.
    – user430825
    Jun 13 '14 at 1:37












  • Additional info: The only useful thing I could derive is that the function has to be either $f(x) = f(-x)$ or $f(x) = -f(-x)$ for all x in A, very easy to establish if one notices that $f(x^2+y) = f((-x)^2+y)$ and setting y = 1. I'd like to discard the possibility of f(x) = f(-x).
    – user430825
    Jun 13 '14 at 21:19


















8














This problem was taken from the bulgarian selection team test for the 47th IMO and appeared in a chinese magazine, I came across it in my own training.



http://www.math.ust.hk/excalibur/v10_n4.pdf



The problem is as follows:



Consider $f:Arightarrow A$ where $A$ is the set of non-zero real numbers. Find all such functions that satisfy the following functional equation for all $x$ and $y$ in $A$:



$$f(x^{2}+y) = f(x)^{2} + frac{f(xy)}{f(x)}$$



I suspect the only function is the identity, but the problem has proved itself to be too difficult for me, I've pondered about it for various hours. Any hint or solution is welcome! Thanks.










share|cite|improve this question
























  • wait...so you are suggesting $f(x) = a^2 + 1$, i.e. f(x) is a constant function? that doesn't work... $f(x^2+y)=a^2+1$ but $f(x)^2 + frac{f(xy)}{f(x)} = a^4 + 2a^2 + 2$
    – user430825
    Jun 13 '14 at 1:20












  • ah I see! thanks for clarification. Just one problem I see, the original formulation $alpha = alpha^{2} + 1$ only has complex solutions, which are outside of the codomain.
    – user430825
    Jun 13 '14 at 1:37












  • Additional info: The only useful thing I could derive is that the function has to be either $f(x) = f(-x)$ or $f(x) = -f(-x)$ for all x in A, very easy to establish if one notices that $f(x^2+y) = f((-x)^2+y)$ and setting y = 1. I'd like to discard the possibility of f(x) = f(-x).
    – user430825
    Jun 13 '14 at 21:19
















8












8








8


3





This problem was taken from the bulgarian selection team test for the 47th IMO and appeared in a chinese magazine, I came across it in my own training.



http://www.math.ust.hk/excalibur/v10_n4.pdf



The problem is as follows:



Consider $f:Arightarrow A$ where $A$ is the set of non-zero real numbers. Find all such functions that satisfy the following functional equation for all $x$ and $y$ in $A$:



$$f(x^{2}+y) = f(x)^{2} + frac{f(xy)}{f(x)}$$



I suspect the only function is the identity, but the problem has proved itself to be too difficult for me, I've pondered about it for various hours. Any hint or solution is welcome! Thanks.










share|cite|improve this question















This problem was taken from the bulgarian selection team test for the 47th IMO and appeared in a chinese magazine, I came across it in my own training.



http://www.math.ust.hk/excalibur/v10_n4.pdf



The problem is as follows:



Consider $f:Arightarrow A$ where $A$ is the set of non-zero real numbers. Find all such functions that satisfy the following functional equation for all $x$ and $y$ in $A$:



$$f(x^{2}+y) = f(x)^{2} + frac{f(xy)}{f(x)}$$



I suspect the only function is the identity, but the problem has proved itself to be too difficult for me, I've pondered about it for various hours. Any hint or solution is welcome! Thanks.







contest-math functional-equations






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edited Jun 14 '14 at 16:24









Daniel Robert-Nicoud

20.3k33596




20.3k33596










asked Jun 12 '14 at 20:13









user430825user430825

412




412












  • wait...so you are suggesting $f(x) = a^2 + 1$, i.e. f(x) is a constant function? that doesn't work... $f(x^2+y)=a^2+1$ but $f(x)^2 + frac{f(xy)}{f(x)} = a^4 + 2a^2 + 2$
    – user430825
    Jun 13 '14 at 1:20












  • ah I see! thanks for clarification. Just one problem I see, the original formulation $alpha = alpha^{2} + 1$ only has complex solutions, which are outside of the codomain.
    – user430825
    Jun 13 '14 at 1:37












  • Additional info: The only useful thing I could derive is that the function has to be either $f(x) = f(-x)$ or $f(x) = -f(-x)$ for all x in A, very easy to establish if one notices that $f(x^2+y) = f((-x)^2+y)$ and setting y = 1. I'd like to discard the possibility of f(x) = f(-x).
    – user430825
    Jun 13 '14 at 21:19




















  • wait...so you are suggesting $f(x) = a^2 + 1$, i.e. f(x) is a constant function? that doesn't work... $f(x^2+y)=a^2+1$ but $f(x)^2 + frac{f(xy)}{f(x)} = a^4 + 2a^2 + 2$
    – user430825
    Jun 13 '14 at 1:20












  • ah I see! thanks for clarification. Just one problem I see, the original formulation $alpha = alpha^{2} + 1$ only has complex solutions, which are outside of the codomain.
    – user430825
    Jun 13 '14 at 1:37












  • Additional info: The only useful thing I could derive is that the function has to be either $f(x) = f(-x)$ or $f(x) = -f(-x)$ for all x in A, very easy to establish if one notices that $f(x^2+y) = f((-x)^2+y)$ and setting y = 1. I'd like to discard the possibility of f(x) = f(-x).
    – user430825
    Jun 13 '14 at 21:19


















wait...so you are suggesting $f(x) = a^2 + 1$, i.e. f(x) is a constant function? that doesn't work... $f(x^2+y)=a^2+1$ but $f(x)^2 + frac{f(xy)}{f(x)} = a^4 + 2a^2 + 2$
– user430825
Jun 13 '14 at 1:20






wait...so you are suggesting $f(x) = a^2 + 1$, i.e. f(x) is a constant function? that doesn't work... $f(x^2+y)=a^2+1$ but $f(x)^2 + frac{f(xy)}{f(x)} = a^4 + 2a^2 + 2$
– user430825
Jun 13 '14 at 1:20














ah I see! thanks for clarification. Just one problem I see, the original formulation $alpha = alpha^{2} + 1$ only has complex solutions, which are outside of the codomain.
– user430825
Jun 13 '14 at 1:37






ah I see! thanks for clarification. Just one problem I see, the original formulation $alpha = alpha^{2} + 1$ only has complex solutions, which are outside of the codomain.
– user430825
Jun 13 '14 at 1:37














Additional info: The only useful thing I could derive is that the function has to be either $f(x) = f(-x)$ or $f(x) = -f(-x)$ for all x in A, very easy to establish if one notices that $f(x^2+y) = f((-x)^2+y)$ and setting y = 1. I'd like to discard the possibility of f(x) = f(-x).
– user430825
Jun 13 '14 at 21:19






Additional info: The only useful thing I could derive is that the function has to be either $f(x) = f(-x)$ or $f(x) = -f(-x)$ for all x in A, very easy to establish if one notices that $f(x^2+y) = f((-x)^2+y)$ and setting y = 1. I'd like to discard the possibility of f(x) = f(-x).
– user430825
Jun 13 '14 at 21:19












3 Answers
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Here is a solution to the actual problem (with a cheap ending).



First because $f(x^2 + y) = f((-x)^2 + y)$, taking $y = 1$ we have
$$ f(x)^2 + 1 = f(-x)^2 + 1 $$
so $f(x)^2 = f(-x)^2$, i.e. $|f(x)| = |f(-x)|$. Thus going back to general $y$,
$$ f(x)^2 + frac{f(xy)}{f(x)} = f(-x)^2 + frac{f(-xy)}{f(-x)} = f(x)^2 + frac{f(- xy)}{f(-x)} $$
hence taking $x = 1$ $$ frac{f(y)}{f(-y)} = frac{f(1)}{f(-1)}. $$
Thus depending on the sign of the rhs, $f$ is either even or odd.



If $f$ is even then $$f(x^2 + y) = f(x)^2 + frac{f(xy)}{f(x)} = f(x)^2 + frac{f(-xy)}{f(x)} = f(x^2 - y).$$
Now for any $z > 1$, let $x^2 = (z + 1)/2$ and $y = (z - 1)/2$. We get that $f(z) = f(1)$. In particular $f(1) = f(2) = f(1^2 + 1) = f(1)^2 + 1$ so $0 = f(1)^2 - f(1) + 1$. Taking the discriminant we see $f(1)$ is not real, contradiction. So $f$ is odd.



Now
$$f(x^2 - y) = f(x)^2 + frac{f(-xy)}{f(x)} = f(x)^2 - frac{f(xy)}{f(x)}$$
so in particular $f(x^2 - 1) = f(x)^2 - 1$. Observe now that since
$$ f(x^2 + 1) = f(x)^2 + 1 $$
we know $f(x) > 1$ for $x > 1$. Furthermore, for $0 < z < 1$ set $z = x^2 - 1$ so $x = sqrt{z + 1} > 1$, hence we know $f(z) = f(x^2 - 1) = f(x)^2 - 1 > 0$. So we have shown that $f$ is positive-valued on $x > 0$. Hence (being lazy) I now appeal to the RELATED PROBLEM to answer the question. Now the intended solution is definitely not as complicated as this (already we know a lot more than in the RELATED PROBLEM), so I encourage you to find your own solution!



BELOW RELATED PROBLEM: as pointed out, below is a solution for $f : mathbb{R}^{>0} to mathbb{R}^{>0}$, with the restriction $x,y > 0$ in the functional equation which is not the original problem. Actually it is probably harder because $y > 0$ is an annoying constraint.



Because $f$ is positive valued, we know that
$$ f(x^2 + y) > f(x)^2 tag{1} $$
and
$$f(x^2 + y) > frac{f(xy)}{f(x)} tag{2}. $$
The following two Observations motivate us to compute $f(1)$ somewhat "analytically":




  1. Taking $y = 1$ gives $f(x^2 + 1) > 1$, i.e. $f(x) > 1$ for $x > 1$.

  2. The solutions to $x^2 + y = x$ are of the form $(x,x - x^2)$ where $x < 1$. Then from (1) we conclude that for any $x < 1$, $f(x) > f(x)^2$ i.e. $f(x) < 1$.


Let's repeat the latter trick, plugging into the actual equation. Then for $x < 1$ we have
$$f(x) = f(x)^2 + frac{f(x^2 - x^3)}{f(x)}$$
and rearranging,
$$ f(x^2 - x^3) = f(x)^2 - f(x)^3. $$



Let $u_0 = 1/2$ (or an arbitrary value below $1$) and inductively let $u_{i + 1} = u_i^2 - u_i^3$. Because $u_{i + 1} < u_i^2 < 1$ and $u_{i + 1} = u_i^2 - u_i^3 > 0$ we have that $lim_{i to infty} u_i = 0$, and a similar argument gives that $lim_{i to infty} f(u_i) = 0$. Now
$$ f(1 + u_i) = f(1)^2 + frac{f(u_i)}{f(1)} $$
and taking the limit
$$ lim_{i to infty} f(1 + u_i) = f(1)^2. $$
Because $1 < f(1 + u_i)$ by Observation 1 we get that $1 le f(1)^2$ hence $1 le f(1)$.



Now go back to the equality and plug in $y = 1$. We get
$$f(x^2 + 1) = f(x)^2 + 1$$
and taking $x = u_i$ and a limit
$$lim_{i to infty} f(u_i^2 + 1) = 1. $$
Taking $x = 1$, $y = u_i^2$ in the equality
$$ f(1 + u_i^2) = f(1)^2 + frac{f(u_i^2)}{f(1)} $$
and taking the limit we get
$$ 1 = lim_{i to infty} left(f(1)^2 + frac{f(u_i^2)}{f(1)}right) $$
so in particular $f(1)^2 le 1$, i.e. $f(1) le 1$.



So finally we have shown $f(1) = 1$. Taking $x = 1$
$$ f(1 + y) = f(1)^2 + frac{f(y)}{f(1)} = 1 + f(y) tag{3}$$
so by induction
$$f(n + r) = n + f(r)$$
for $n in mathbb{Z}^{> 0}$, $r ge 0$.
Taking $y = 1$ in the original equation
$$ f(x^2 + 1) = f(x)^2 + 1 $$
so by (3), for all $x$ $f(x^2) = f(x)^2$, i.e. $f(sqrt{x}) = sqrt{f(x)}$. Now recalling the original equation
$$ f(x^2 + y) = f(x)^2 + frac{f(xy)}{f(x)} = f(x^2) + frac{f(xy)}{f(x)} tag{4} $$
so $f$ is an increasing function. Recalling our sequence $u_i$ with $u_i to 0, f(u_i) to 0$ as $i to infty$, we conclude that $lim_{x to^+ 0} f(x) = 0$. Taking the limit in (4),
$$ lim_{y to^+ 0} f(x^2 + y) = f(x^2) $$
and so $f$ is right-continuous everywhere.
Fix arbitrary $z$, and let $x^2 = z - y$.
Then
$$ f(z) - f(z - y) = f(x^2 + y) - f(x^2) = frac{f(xy)}{f(x)} < frac{f(sqrt{z} cdot y)}{f(sqrt{z - k})} $$
for any fixed $k > 0$ such that $0 < z - k$, for all $y < k$ (so $sqrt{z - k} < sqrt{z - y}$). Taking the limit as $y to^+ 0$ we find
that $f$ is left-continuous at $z$. Hence $f$ is continuous everywhere.



I now use the (nontrivial, but intuitive) fact that non-negative multiples of $sqrt{2}$ are dense in $mathbb{R}/mathbb{Z}$. In other words
$$ {asqrt{2} + b : a in mathbb{Z}^{ge 0},b in mathbb{Z}} $$
is dense in $mathbb{R}$. For any positive $asqrt{2} + b = sqrt{2a^2} + b$, if $b ge 0$ then
we know
$$ f(sqrt{2a^2} + b) = b + f(sqrt{2a^2}) = b + sqrt{2a^2}. $$
If $b < 0$ then we know
$$ f(sqrt{2a^2} + b) - b = f(sqrt{2a^2} + b - b) = sqrt{2a^2}. $$
So $f$ equals the identity on a dense subset of $mathbb{R}^{> 0}$. By continuity, $f = id$.



PS This might not be the shortest solution! At least it is fairly systematic. Also, my apologies, there may be small mistakes, so caveat emptor.






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  • This is a very nice and titanic solution :P but the first assumption, namely that f is positively valued, isn't an hypothesis in the original formulation from what I understand. Anyway there are a couple of very nice methods here that I'm sure are fruitful and will keep me busy revisiting and adapting them. PD: I can't upvote your response because I don't have enough reputation. Thank you VERY much for your time bro.
    – user430825
    Jun 14 '14 at 1:57












  • Oops! I misread the problem. At least you know how to solve the problem when f is positive-valued now. I will think about it when f is on R^*, possibly there is also a shorter solution then, haha.
    – Frederic Koehler
    Jun 14 '14 at 2:10










  • @user430825 I have uploaded a (hopefully correct) solution to the correct problem, but feel free to try to solve it on your own first! the problem was fun, you're welcome.
    – Frederic Koehler
    Jun 14 '14 at 4:54










  • very clever my friend, the part where you show that it cannot be an even function, I was working on that one! hehe. I'll try to come up with a solution that doesn't appeal to your special case! Thanks again for your time.
    – user430825
    Jun 14 '14 at 10:58



















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This might not be the optimal approach... but it attempts to avoid reasoning based on real analysis (well, apart from the very definition of irrational numbers).



Determine the value of $f(1)$



Let $f(1)=A$. Then, for any $y$ different from $0$ and $(-1)$ we have $$f(1+y)=f(1)^2 + f(y)/f(1) = A^2 + f(y)/A$$ which allows us to express the following quantities: $$begin{eqnarray}
f(2) & = & A^2 + 1 \
f(3) & = & A^2 + A + frac{1}{A} \
f(4) & = & A^2 + A + 1 + frac{1}{A^2} \
f(5) & = & A^2 + A + 1 + frac{1}{A} + frac{1}{A^3} \
end{eqnarray}$$



At the same time, we have $f(x^2+1)=f(x)^2+f(x)/f(x)=f(x)^2+1$. This gives us $$f(5)=f(2)^2+1$$ Equating the two expressions for $f(5)$ and substituting for $f(2)$ gives us
$$(A^2+1)^2 + 1 = A^2+A+1+frac{1}{A} + frac{1}{A^3}$$ which simplifies to $$begin{eqnarray}0 & = & A^7 + A^5 - A^4 + A^3 - A^2 - 1 \ & = & (A-1)(A^2-A+1)(A^2+A+1)^2end{eqnarray}$$ whose only real root is $A=1$. Thus, $f(1)=1$ and we have $f(1+y)=1+f(y)$ for all $y$ other than $0$ and $(-1)$. In particular, this means that $f(k)=k$ for any positive integer $k$.



Show additive property for numbers of same sign



Now, consider real numbers $u,v$ of the same sign and these two equalities:
$$begin{eqnarray}
f(v^2 + u/v) & = & f(v)^2 + f(u)/f(v) \
f(v^2 + (u/v + 1)) & = & f(v)^2 + f(u+v)/f(v) \
end{eqnarray}$$
We can apply the equality $f(1+y)=1+f(y)$ since both $v^2$ and $u/v$ are positive to obtain
$$f(u+v)=f(u)+f(v)$$



Note: This equation is very similar to Cauchy's functional equation; a quite interesting problem of its own. The equality given in our problem is stronger than "just" plain additivity, though.



Determine values of $f$ on positive rationals



It's quite easy to see that this equality implies that $fleft(frac{p}{q}right)=frac{p}{q}$ for any positive integers $p,q$; one simply expands $$p=f(p)=fleft(frac{p}{q}right)+fleft(frac{p}{q}right)+ldots+fleft(frac{p}{q}right)=qcdot fleft(frac{p}{q}right)$$



Show that $f$ is increasing on positive reals



For $y>0$, we can write $$begin{eqnarray}
f(y) = f(y)+1-1 & = & f(y+1)-1 \
& = & fleft((sqrt{y})^2+1right)-1 = \
& = & fleft(sqrt{y}right)^2+1-1 \
& = & fleft(sqrt{y}right)^2>0
end{eqnarray}$$



Thus, for any positive reals $u>v>0$, we get $$f(u)=f(v+(u-v))=f(v)+f(u-v)>f(v)$$



Determine the values of $f$ for positive reals



Let's show that $f(x)=x$ for $x>0$. If that wasn't the case, we'd have either $f(x)>x$ or $0<f(x)<x$. In the first case, we could find rational number $z$ with $f(x)>z>x$. First inequality implies $f(x)>f(z)$ since $f(z)=z$, while the second one implies $f(z)>f(x)$, a contradiction. The other case can be treated in the same way. Thus, $f(x)=x$.



Alternate approach is to realize that the Dedekind cut corresponding to irrational $x>0$ must be the same as the Dedekind cut corresponding to $f(x)$.



Finish the proof by considering the negative reals



If $x<0$ is non-integer, we can simply use the equality $f(1+y)=1+f(y)$ to show that $f(x)=x$; just by adding $1$ until we reach a positive number. For negative integers, we can bootstrap the process by considering $f(-1)=fleft(-frac{1}{2}right)+fleft(-frac{1}{2}right)=(-1)$ and do the same by reaching $(-1)$.



The only function satisfying the given condition is $f(x)=x$ Q.E.D.






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  • Excellent answer. Really nice. Thanks for your time!
    – user430825
    Jun 14 '14 at 17:22





















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$newcommand{NN}{{mathbb{N}}}newcommand{CC}{{mathbb{C}}}newcommand{RR}{{mathbb{R}}}newcommand{QQ}{{mathbb Q}}newcommand{ra}{rightarrow}newcommand{ds}{displaystyle}newcommand{mat}[1]{left(begin{matrix}#1end{matrix}right)}newcommand{lam}{lambda}renewcommand{P}{{mathcal P}}newcommand{N}{{mathcal N}}newcommand{M}{{mathcal M}}renewcommand{L}{{mathcal L}}newcommand{EQ}[2]{begin{equation} {{#2}label{#1}} end{equation}}newcommand{eps}{varepsilon}$
It contains numerous elements that are also in other answers - I put it anyway.
Observe that $f$ cannot be constant, because the equation $C=C^2+1$ has no real solution.



Putting $y=1$, we find that $f(x)^2+1=f(x^2+1)=f(-x)^2+1$ and hence $f(x)^2=f(-x)^2$
for all $xin A$. Putting $x=pm1$, we obtain $f(-y)/f(-1)=f(y)/f(1)$ for all $y$. So either
$f(-x)=f(x)$ for all $x$ or $f(-x)=-f(x)$ for all $x$.



The solution $y$ of $x^2+y=xy$ is $y=frac{x^2}{x-1}$ unless $x=1$. Using this $y$ we have
$$fleft(frac{x^3}{x-1}right)=frac{f(x)^3}{f(x)-1}mbox{ for }xneq1.$$
Consider now a solution $z$ of $frac{z^2}{z-1}=-1$. Then $f(-z)=frac{f(z)^3}{f(z)-1}$.
As a consequence $f(-z)$ cannot equal $f(z)$ because $1=frac{f(z)^2}{f(z)-1}$ has no real
solution for $f(z)$. Hence $f(-z)=-f(z)$ and, as seen above, $f(-x)=-f(x)$ for all $x.$



Next consider $x=1$ and replace $y$ by $-y$. We find $f(1-y)=f(1)^2-frac{f(y)}{f(1)}$
for $yneq1$. Replacing here $y$ by $1-y$ yields with $w=f(1)$
$$w^2f(y)=w^4-wf(1-y)=w^4-w^3+f(y)mbox{ for all }yin A,yneq1.$$
Thus $(w^2-1)f(y)=w^4-w^3$. So we must have $w=1$ as otherwise $w=-1$
which leads to a contradiction or $f(y)=frac{w^3}{w+1}$ is constant which had been excluded.
So finally we proved that $w=f(1)=1$.



Putting $x=1$ now gives $f(y+1)=f(y)+1$ for $yneq-1$. In particular, we find $f(n)=n$ and
$f(-n)=-n$ for all positive integers $n$. Putting $x=n$, we obtain
$f(y+n^2)=n^2+frac1nf(ny)$ for all $yneq-n^2$ and $f(ny)=nf(y)$ for all positive $y$
(In fact, we only have to exclude that $y$ is a negative integer).
In particular this leads to $f(q)=q$ for all rational positive $q$. It is extended to
negative rationals using $f(-x)=-f(x)$.
Moreover, we obtain $f(y+m)=f(y)+m$ for all non-integer $y$ and thus
$f(y+frac mn)=frac1nf(ny+m)=frac1n(nf(y)+m)=f(y)+frac mn$
for all irrational $y$ and hence $f(y+q)=f(y)+q$
for all rational $q$ and irrational $y$.



Finally, as $f(x^2+1)=f(x)^2+1$ and hence $f(x^2)=f(x)^2$ for all $x$,
the values of $f$ for positive $y$ are positive and hence also its values for
negative $y$ are negative.
Given any irrational $y$, we consider rational $q,r$ with $q<y<r$. Then
as $y-q>0$, we have $f(y)-q=f(y-q)>0$ and hence $q<f(y)$. In the same way we show that
$f(y)<r$. Since we can use any rational $q,r$ with $q<y<r$,
we have shown that $f(y)=y$ also for irrational $y$.






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    Here is a solution to the actual problem (with a cheap ending).



    First because $f(x^2 + y) = f((-x)^2 + y)$, taking $y = 1$ we have
    $$ f(x)^2 + 1 = f(-x)^2 + 1 $$
    so $f(x)^2 = f(-x)^2$, i.e. $|f(x)| = |f(-x)|$. Thus going back to general $y$,
    $$ f(x)^2 + frac{f(xy)}{f(x)} = f(-x)^2 + frac{f(-xy)}{f(-x)} = f(x)^2 + frac{f(- xy)}{f(-x)} $$
    hence taking $x = 1$ $$ frac{f(y)}{f(-y)} = frac{f(1)}{f(-1)}. $$
    Thus depending on the sign of the rhs, $f$ is either even or odd.



    If $f$ is even then $$f(x^2 + y) = f(x)^2 + frac{f(xy)}{f(x)} = f(x)^2 + frac{f(-xy)}{f(x)} = f(x^2 - y).$$
    Now for any $z > 1$, let $x^2 = (z + 1)/2$ and $y = (z - 1)/2$. We get that $f(z) = f(1)$. In particular $f(1) = f(2) = f(1^2 + 1) = f(1)^2 + 1$ so $0 = f(1)^2 - f(1) + 1$. Taking the discriminant we see $f(1)$ is not real, contradiction. So $f$ is odd.



    Now
    $$f(x^2 - y) = f(x)^2 + frac{f(-xy)}{f(x)} = f(x)^2 - frac{f(xy)}{f(x)}$$
    so in particular $f(x^2 - 1) = f(x)^2 - 1$. Observe now that since
    $$ f(x^2 + 1) = f(x)^2 + 1 $$
    we know $f(x) > 1$ for $x > 1$. Furthermore, for $0 < z < 1$ set $z = x^2 - 1$ so $x = sqrt{z + 1} > 1$, hence we know $f(z) = f(x^2 - 1) = f(x)^2 - 1 > 0$. So we have shown that $f$ is positive-valued on $x > 0$. Hence (being lazy) I now appeal to the RELATED PROBLEM to answer the question. Now the intended solution is definitely not as complicated as this (already we know a lot more than in the RELATED PROBLEM), so I encourage you to find your own solution!



    BELOW RELATED PROBLEM: as pointed out, below is a solution for $f : mathbb{R}^{>0} to mathbb{R}^{>0}$, with the restriction $x,y > 0$ in the functional equation which is not the original problem. Actually it is probably harder because $y > 0$ is an annoying constraint.



    Because $f$ is positive valued, we know that
    $$ f(x^2 + y) > f(x)^2 tag{1} $$
    and
    $$f(x^2 + y) > frac{f(xy)}{f(x)} tag{2}. $$
    The following two Observations motivate us to compute $f(1)$ somewhat "analytically":




    1. Taking $y = 1$ gives $f(x^2 + 1) > 1$, i.e. $f(x) > 1$ for $x > 1$.

    2. The solutions to $x^2 + y = x$ are of the form $(x,x - x^2)$ where $x < 1$. Then from (1) we conclude that for any $x < 1$, $f(x) > f(x)^2$ i.e. $f(x) < 1$.


    Let's repeat the latter trick, plugging into the actual equation. Then for $x < 1$ we have
    $$f(x) = f(x)^2 + frac{f(x^2 - x^3)}{f(x)}$$
    and rearranging,
    $$ f(x^2 - x^3) = f(x)^2 - f(x)^3. $$



    Let $u_0 = 1/2$ (or an arbitrary value below $1$) and inductively let $u_{i + 1} = u_i^2 - u_i^3$. Because $u_{i + 1} < u_i^2 < 1$ and $u_{i + 1} = u_i^2 - u_i^3 > 0$ we have that $lim_{i to infty} u_i = 0$, and a similar argument gives that $lim_{i to infty} f(u_i) = 0$. Now
    $$ f(1 + u_i) = f(1)^2 + frac{f(u_i)}{f(1)} $$
    and taking the limit
    $$ lim_{i to infty} f(1 + u_i) = f(1)^2. $$
    Because $1 < f(1 + u_i)$ by Observation 1 we get that $1 le f(1)^2$ hence $1 le f(1)$.



    Now go back to the equality and plug in $y = 1$. We get
    $$f(x^2 + 1) = f(x)^2 + 1$$
    and taking $x = u_i$ and a limit
    $$lim_{i to infty} f(u_i^2 + 1) = 1. $$
    Taking $x = 1$, $y = u_i^2$ in the equality
    $$ f(1 + u_i^2) = f(1)^2 + frac{f(u_i^2)}{f(1)} $$
    and taking the limit we get
    $$ 1 = lim_{i to infty} left(f(1)^2 + frac{f(u_i^2)}{f(1)}right) $$
    so in particular $f(1)^2 le 1$, i.e. $f(1) le 1$.



    So finally we have shown $f(1) = 1$. Taking $x = 1$
    $$ f(1 + y) = f(1)^2 + frac{f(y)}{f(1)} = 1 + f(y) tag{3}$$
    so by induction
    $$f(n + r) = n + f(r)$$
    for $n in mathbb{Z}^{> 0}$, $r ge 0$.
    Taking $y = 1$ in the original equation
    $$ f(x^2 + 1) = f(x)^2 + 1 $$
    so by (3), for all $x$ $f(x^2) = f(x)^2$, i.e. $f(sqrt{x}) = sqrt{f(x)}$. Now recalling the original equation
    $$ f(x^2 + y) = f(x)^2 + frac{f(xy)}{f(x)} = f(x^2) + frac{f(xy)}{f(x)} tag{4} $$
    so $f$ is an increasing function. Recalling our sequence $u_i$ with $u_i to 0, f(u_i) to 0$ as $i to infty$, we conclude that $lim_{x to^+ 0} f(x) = 0$. Taking the limit in (4),
    $$ lim_{y to^+ 0} f(x^2 + y) = f(x^2) $$
    and so $f$ is right-continuous everywhere.
    Fix arbitrary $z$, and let $x^2 = z - y$.
    Then
    $$ f(z) - f(z - y) = f(x^2 + y) - f(x^2) = frac{f(xy)}{f(x)} < frac{f(sqrt{z} cdot y)}{f(sqrt{z - k})} $$
    for any fixed $k > 0$ such that $0 < z - k$, for all $y < k$ (so $sqrt{z - k} < sqrt{z - y}$). Taking the limit as $y to^+ 0$ we find
    that $f$ is left-continuous at $z$. Hence $f$ is continuous everywhere.



    I now use the (nontrivial, but intuitive) fact that non-negative multiples of $sqrt{2}$ are dense in $mathbb{R}/mathbb{Z}$. In other words
    $$ {asqrt{2} + b : a in mathbb{Z}^{ge 0},b in mathbb{Z}} $$
    is dense in $mathbb{R}$. For any positive $asqrt{2} + b = sqrt{2a^2} + b$, if $b ge 0$ then
    we know
    $$ f(sqrt{2a^2} + b) = b + f(sqrt{2a^2}) = b + sqrt{2a^2}. $$
    If $b < 0$ then we know
    $$ f(sqrt{2a^2} + b) - b = f(sqrt{2a^2} + b - b) = sqrt{2a^2}. $$
    So $f$ equals the identity on a dense subset of $mathbb{R}^{> 0}$. By continuity, $f = id$.



    PS This might not be the shortest solution! At least it is fairly systematic. Also, my apologies, there may be small mistakes, so caveat emptor.






    share|cite|improve this answer























    • This is a very nice and titanic solution :P but the first assumption, namely that f is positively valued, isn't an hypothesis in the original formulation from what I understand. Anyway there are a couple of very nice methods here that I'm sure are fruitful and will keep me busy revisiting and adapting them. PD: I can't upvote your response because I don't have enough reputation. Thank you VERY much for your time bro.
      – user430825
      Jun 14 '14 at 1:57












    • Oops! I misread the problem. At least you know how to solve the problem when f is positive-valued now. I will think about it when f is on R^*, possibly there is also a shorter solution then, haha.
      – Frederic Koehler
      Jun 14 '14 at 2:10










    • @user430825 I have uploaded a (hopefully correct) solution to the correct problem, but feel free to try to solve it on your own first! the problem was fun, you're welcome.
      – Frederic Koehler
      Jun 14 '14 at 4:54










    • very clever my friend, the part where you show that it cannot be an even function, I was working on that one! hehe. I'll try to come up with a solution that doesn't appeal to your special case! Thanks again for your time.
      – user430825
      Jun 14 '14 at 10:58
















    5














    Here is a solution to the actual problem (with a cheap ending).



    First because $f(x^2 + y) = f((-x)^2 + y)$, taking $y = 1$ we have
    $$ f(x)^2 + 1 = f(-x)^2 + 1 $$
    so $f(x)^2 = f(-x)^2$, i.e. $|f(x)| = |f(-x)|$. Thus going back to general $y$,
    $$ f(x)^2 + frac{f(xy)}{f(x)} = f(-x)^2 + frac{f(-xy)}{f(-x)} = f(x)^2 + frac{f(- xy)}{f(-x)} $$
    hence taking $x = 1$ $$ frac{f(y)}{f(-y)} = frac{f(1)}{f(-1)}. $$
    Thus depending on the sign of the rhs, $f$ is either even or odd.



    If $f$ is even then $$f(x^2 + y) = f(x)^2 + frac{f(xy)}{f(x)} = f(x)^2 + frac{f(-xy)}{f(x)} = f(x^2 - y).$$
    Now for any $z > 1$, let $x^2 = (z + 1)/2$ and $y = (z - 1)/2$. We get that $f(z) = f(1)$. In particular $f(1) = f(2) = f(1^2 + 1) = f(1)^2 + 1$ so $0 = f(1)^2 - f(1) + 1$. Taking the discriminant we see $f(1)$ is not real, contradiction. So $f$ is odd.



    Now
    $$f(x^2 - y) = f(x)^2 + frac{f(-xy)}{f(x)} = f(x)^2 - frac{f(xy)}{f(x)}$$
    so in particular $f(x^2 - 1) = f(x)^2 - 1$. Observe now that since
    $$ f(x^2 + 1) = f(x)^2 + 1 $$
    we know $f(x) > 1$ for $x > 1$. Furthermore, for $0 < z < 1$ set $z = x^2 - 1$ so $x = sqrt{z + 1} > 1$, hence we know $f(z) = f(x^2 - 1) = f(x)^2 - 1 > 0$. So we have shown that $f$ is positive-valued on $x > 0$. Hence (being lazy) I now appeal to the RELATED PROBLEM to answer the question. Now the intended solution is definitely not as complicated as this (already we know a lot more than in the RELATED PROBLEM), so I encourage you to find your own solution!



    BELOW RELATED PROBLEM: as pointed out, below is a solution for $f : mathbb{R}^{>0} to mathbb{R}^{>0}$, with the restriction $x,y > 0$ in the functional equation which is not the original problem. Actually it is probably harder because $y > 0$ is an annoying constraint.



    Because $f$ is positive valued, we know that
    $$ f(x^2 + y) > f(x)^2 tag{1} $$
    and
    $$f(x^2 + y) > frac{f(xy)}{f(x)} tag{2}. $$
    The following two Observations motivate us to compute $f(1)$ somewhat "analytically":




    1. Taking $y = 1$ gives $f(x^2 + 1) > 1$, i.e. $f(x) > 1$ for $x > 1$.

    2. The solutions to $x^2 + y = x$ are of the form $(x,x - x^2)$ where $x < 1$. Then from (1) we conclude that for any $x < 1$, $f(x) > f(x)^2$ i.e. $f(x) < 1$.


    Let's repeat the latter trick, plugging into the actual equation. Then for $x < 1$ we have
    $$f(x) = f(x)^2 + frac{f(x^2 - x^3)}{f(x)}$$
    and rearranging,
    $$ f(x^2 - x^3) = f(x)^2 - f(x)^3. $$



    Let $u_0 = 1/2$ (or an arbitrary value below $1$) and inductively let $u_{i + 1} = u_i^2 - u_i^3$. Because $u_{i + 1} < u_i^2 < 1$ and $u_{i + 1} = u_i^2 - u_i^3 > 0$ we have that $lim_{i to infty} u_i = 0$, and a similar argument gives that $lim_{i to infty} f(u_i) = 0$. Now
    $$ f(1 + u_i) = f(1)^2 + frac{f(u_i)}{f(1)} $$
    and taking the limit
    $$ lim_{i to infty} f(1 + u_i) = f(1)^2. $$
    Because $1 < f(1 + u_i)$ by Observation 1 we get that $1 le f(1)^2$ hence $1 le f(1)$.



    Now go back to the equality and plug in $y = 1$. We get
    $$f(x^2 + 1) = f(x)^2 + 1$$
    and taking $x = u_i$ and a limit
    $$lim_{i to infty} f(u_i^2 + 1) = 1. $$
    Taking $x = 1$, $y = u_i^2$ in the equality
    $$ f(1 + u_i^2) = f(1)^2 + frac{f(u_i^2)}{f(1)} $$
    and taking the limit we get
    $$ 1 = lim_{i to infty} left(f(1)^2 + frac{f(u_i^2)}{f(1)}right) $$
    so in particular $f(1)^2 le 1$, i.e. $f(1) le 1$.



    So finally we have shown $f(1) = 1$. Taking $x = 1$
    $$ f(1 + y) = f(1)^2 + frac{f(y)}{f(1)} = 1 + f(y) tag{3}$$
    so by induction
    $$f(n + r) = n + f(r)$$
    for $n in mathbb{Z}^{> 0}$, $r ge 0$.
    Taking $y = 1$ in the original equation
    $$ f(x^2 + 1) = f(x)^2 + 1 $$
    so by (3), for all $x$ $f(x^2) = f(x)^2$, i.e. $f(sqrt{x}) = sqrt{f(x)}$. Now recalling the original equation
    $$ f(x^2 + y) = f(x)^2 + frac{f(xy)}{f(x)} = f(x^2) + frac{f(xy)}{f(x)} tag{4} $$
    so $f$ is an increasing function. Recalling our sequence $u_i$ with $u_i to 0, f(u_i) to 0$ as $i to infty$, we conclude that $lim_{x to^+ 0} f(x) = 0$. Taking the limit in (4),
    $$ lim_{y to^+ 0} f(x^2 + y) = f(x^2) $$
    and so $f$ is right-continuous everywhere.
    Fix arbitrary $z$, and let $x^2 = z - y$.
    Then
    $$ f(z) - f(z - y) = f(x^2 + y) - f(x^2) = frac{f(xy)}{f(x)} < frac{f(sqrt{z} cdot y)}{f(sqrt{z - k})} $$
    for any fixed $k > 0$ such that $0 < z - k$, for all $y < k$ (so $sqrt{z - k} < sqrt{z - y}$). Taking the limit as $y to^+ 0$ we find
    that $f$ is left-continuous at $z$. Hence $f$ is continuous everywhere.



    I now use the (nontrivial, but intuitive) fact that non-negative multiples of $sqrt{2}$ are dense in $mathbb{R}/mathbb{Z}$. In other words
    $$ {asqrt{2} + b : a in mathbb{Z}^{ge 0},b in mathbb{Z}} $$
    is dense in $mathbb{R}$. For any positive $asqrt{2} + b = sqrt{2a^2} + b$, if $b ge 0$ then
    we know
    $$ f(sqrt{2a^2} + b) = b + f(sqrt{2a^2}) = b + sqrt{2a^2}. $$
    If $b < 0$ then we know
    $$ f(sqrt{2a^2} + b) - b = f(sqrt{2a^2} + b - b) = sqrt{2a^2}. $$
    So $f$ equals the identity on a dense subset of $mathbb{R}^{> 0}$. By continuity, $f = id$.



    PS This might not be the shortest solution! At least it is fairly systematic. Also, my apologies, there may be small mistakes, so caveat emptor.






    share|cite|improve this answer























    • This is a very nice and titanic solution :P but the first assumption, namely that f is positively valued, isn't an hypothesis in the original formulation from what I understand. Anyway there are a couple of very nice methods here that I'm sure are fruitful and will keep me busy revisiting and adapting them. PD: I can't upvote your response because I don't have enough reputation. Thank you VERY much for your time bro.
      – user430825
      Jun 14 '14 at 1:57












    • Oops! I misread the problem. At least you know how to solve the problem when f is positive-valued now. I will think about it when f is on R^*, possibly there is also a shorter solution then, haha.
      – Frederic Koehler
      Jun 14 '14 at 2:10










    • @user430825 I have uploaded a (hopefully correct) solution to the correct problem, but feel free to try to solve it on your own first! the problem was fun, you're welcome.
      – Frederic Koehler
      Jun 14 '14 at 4:54










    • very clever my friend, the part where you show that it cannot be an even function, I was working on that one! hehe. I'll try to come up with a solution that doesn't appeal to your special case! Thanks again for your time.
      – user430825
      Jun 14 '14 at 10:58














    5












    5








    5






    Here is a solution to the actual problem (with a cheap ending).



    First because $f(x^2 + y) = f((-x)^2 + y)$, taking $y = 1$ we have
    $$ f(x)^2 + 1 = f(-x)^2 + 1 $$
    so $f(x)^2 = f(-x)^2$, i.e. $|f(x)| = |f(-x)|$. Thus going back to general $y$,
    $$ f(x)^2 + frac{f(xy)}{f(x)} = f(-x)^2 + frac{f(-xy)}{f(-x)} = f(x)^2 + frac{f(- xy)}{f(-x)} $$
    hence taking $x = 1$ $$ frac{f(y)}{f(-y)} = frac{f(1)}{f(-1)}. $$
    Thus depending on the sign of the rhs, $f$ is either even or odd.



    If $f$ is even then $$f(x^2 + y) = f(x)^2 + frac{f(xy)}{f(x)} = f(x)^2 + frac{f(-xy)}{f(x)} = f(x^2 - y).$$
    Now for any $z > 1$, let $x^2 = (z + 1)/2$ and $y = (z - 1)/2$. We get that $f(z) = f(1)$. In particular $f(1) = f(2) = f(1^2 + 1) = f(1)^2 + 1$ so $0 = f(1)^2 - f(1) + 1$. Taking the discriminant we see $f(1)$ is not real, contradiction. So $f$ is odd.



    Now
    $$f(x^2 - y) = f(x)^2 + frac{f(-xy)}{f(x)} = f(x)^2 - frac{f(xy)}{f(x)}$$
    so in particular $f(x^2 - 1) = f(x)^2 - 1$. Observe now that since
    $$ f(x^2 + 1) = f(x)^2 + 1 $$
    we know $f(x) > 1$ for $x > 1$. Furthermore, for $0 < z < 1$ set $z = x^2 - 1$ so $x = sqrt{z + 1} > 1$, hence we know $f(z) = f(x^2 - 1) = f(x)^2 - 1 > 0$. So we have shown that $f$ is positive-valued on $x > 0$. Hence (being lazy) I now appeal to the RELATED PROBLEM to answer the question. Now the intended solution is definitely not as complicated as this (already we know a lot more than in the RELATED PROBLEM), so I encourage you to find your own solution!



    BELOW RELATED PROBLEM: as pointed out, below is a solution for $f : mathbb{R}^{>0} to mathbb{R}^{>0}$, with the restriction $x,y > 0$ in the functional equation which is not the original problem. Actually it is probably harder because $y > 0$ is an annoying constraint.



    Because $f$ is positive valued, we know that
    $$ f(x^2 + y) > f(x)^2 tag{1} $$
    and
    $$f(x^2 + y) > frac{f(xy)}{f(x)} tag{2}. $$
    The following two Observations motivate us to compute $f(1)$ somewhat "analytically":




    1. Taking $y = 1$ gives $f(x^2 + 1) > 1$, i.e. $f(x) > 1$ for $x > 1$.

    2. The solutions to $x^2 + y = x$ are of the form $(x,x - x^2)$ where $x < 1$. Then from (1) we conclude that for any $x < 1$, $f(x) > f(x)^2$ i.e. $f(x) < 1$.


    Let's repeat the latter trick, plugging into the actual equation. Then for $x < 1$ we have
    $$f(x) = f(x)^2 + frac{f(x^2 - x^3)}{f(x)}$$
    and rearranging,
    $$ f(x^2 - x^3) = f(x)^2 - f(x)^3. $$



    Let $u_0 = 1/2$ (or an arbitrary value below $1$) and inductively let $u_{i + 1} = u_i^2 - u_i^3$. Because $u_{i + 1} < u_i^2 < 1$ and $u_{i + 1} = u_i^2 - u_i^3 > 0$ we have that $lim_{i to infty} u_i = 0$, and a similar argument gives that $lim_{i to infty} f(u_i) = 0$. Now
    $$ f(1 + u_i) = f(1)^2 + frac{f(u_i)}{f(1)} $$
    and taking the limit
    $$ lim_{i to infty} f(1 + u_i) = f(1)^2. $$
    Because $1 < f(1 + u_i)$ by Observation 1 we get that $1 le f(1)^2$ hence $1 le f(1)$.



    Now go back to the equality and plug in $y = 1$. We get
    $$f(x^2 + 1) = f(x)^2 + 1$$
    and taking $x = u_i$ and a limit
    $$lim_{i to infty} f(u_i^2 + 1) = 1. $$
    Taking $x = 1$, $y = u_i^2$ in the equality
    $$ f(1 + u_i^2) = f(1)^2 + frac{f(u_i^2)}{f(1)} $$
    and taking the limit we get
    $$ 1 = lim_{i to infty} left(f(1)^2 + frac{f(u_i^2)}{f(1)}right) $$
    so in particular $f(1)^2 le 1$, i.e. $f(1) le 1$.



    So finally we have shown $f(1) = 1$. Taking $x = 1$
    $$ f(1 + y) = f(1)^2 + frac{f(y)}{f(1)} = 1 + f(y) tag{3}$$
    so by induction
    $$f(n + r) = n + f(r)$$
    for $n in mathbb{Z}^{> 0}$, $r ge 0$.
    Taking $y = 1$ in the original equation
    $$ f(x^2 + 1) = f(x)^2 + 1 $$
    so by (3), for all $x$ $f(x^2) = f(x)^2$, i.e. $f(sqrt{x}) = sqrt{f(x)}$. Now recalling the original equation
    $$ f(x^2 + y) = f(x)^2 + frac{f(xy)}{f(x)} = f(x^2) + frac{f(xy)}{f(x)} tag{4} $$
    so $f$ is an increasing function. Recalling our sequence $u_i$ with $u_i to 0, f(u_i) to 0$ as $i to infty$, we conclude that $lim_{x to^+ 0} f(x) = 0$. Taking the limit in (4),
    $$ lim_{y to^+ 0} f(x^2 + y) = f(x^2) $$
    and so $f$ is right-continuous everywhere.
    Fix arbitrary $z$, and let $x^2 = z - y$.
    Then
    $$ f(z) - f(z - y) = f(x^2 + y) - f(x^2) = frac{f(xy)}{f(x)} < frac{f(sqrt{z} cdot y)}{f(sqrt{z - k})} $$
    for any fixed $k > 0$ such that $0 < z - k$, for all $y < k$ (so $sqrt{z - k} < sqrt{z - y}$). Taking the limit as $y to^+ 0$ we find
    that $f$ is left-continuous at $z$. Hence $f$ is continuous everywhere.



    I now use the (nontrivial, but intuitive) fact that non-negative multiples of $sqrt{2}$ are dense in $mathbb{R}/mathbb{Z}$. In other words
    $$ {asqrt{2} + b : a in mathbb{Z}^{ge 0},b in mathbb{Z}} $$
    is dense in $mathbb{R}$. For any positive $asqrt{2} + b = sqrt{2a^2} + b$, if $b ge 0$ then
    we know
    $$ f(sqrt{2a^2} + b) = b + f(sqrt{2a^2}) = b + sqrt{2a^2}. $$
    If $b < 0$ then we know
    $$ f(sqrt{2a^2} + b) - b = f(sqrt{2a^2} + b - b) = sqrt{2a^2}. $$
    So $f$ equals the identity on a dense subset of $mathbb{R}^{> 0}$. By continuity, $f = id$.



    PS This might not be the shortest solution! At least it is fairly systematic. Also, my apologies, there may be small mistakes, so caveat emptor.






    share|cite|improve this answer














    Here is a solution to the actual problem (with a cheap ending).



    First because $f(x^2 + y) = f((-x)^2 + y)$, taking $y = 1$ we have
    $$ f(x)^2 + 1 = f(-x)^2 + 1 $$
    so $f(x)^2 = f(-x)^2$, i.e. $|f(x)| = |f(-x)|$. Thus going back to general $y$,
    $$ f(x)^2 + frac{f(xy)}{f(x)} = f(-x)^2 + frac{f(-xy)}{f(-x)} = f(x)^2 + frac{f(- xy)}{f(-x)} $$
    hence taking $x = 1$ $$ frac{f(y)}{f(-y)} = frac{f(1)}{f(-1)}. $$
    Thus depending on the sign of the rhs, $f$ is either even or odd.



    If $f$ is even then $$f(x^2 + y) = f(x)^2 + frac{f(xy)}{f(x)} = f(x)^2 + frac{f(-xy)}{f(x)} = f(x^2 - y).$$
    Now for any $z > 1$, let $x^2 = (z + 1)/2$ and $y = (z - 1)/2$. We get that $f(z) = f(1)$. In particular $f(1) = f(2) = f(1^2 + 1) = f(1)^2 + 1$ so $0 = f(1)^2 - f(1) + 1$. Taking the discriminant we see $f(1)$ is not real, contradiction. So $f$ is odd.



    Now
    $$f(x^2 - y) = f(x)^2 + frac{f(-xy)}{f(x)} = f(x)^2 - frac{f(xy)}{f(x)}$$
    so in particular $f(x^2 - 1) = f(x)^2 - 1$. Observe now that since
    $$ f(x^2 + 1) = f(x)^2 + 1 $$
    we know $f(x) > 1$ for $x > 1$. Furthermore, for $0 < z < 1$ set $z = x^2 - 1$ so $x = sqrt{z + 1} > 1$, hence we know $f(z) = f(x^2 - 1) = f(x)^2 - 1 > 0$. So we have shown that $f$ is positive-valued on $x > 0$. Hence (being lazy) I now appeal to the RELATED PROBLEM to answer the question. Now the intended solution is definitely not as complicated as this (already we know a lot more than in the RELATED PROBLEM), so I encourage you to find your own solution!



    BELOW RELATED PROBLEM: as pointed out, below is a solution for $f : mathbb{R}^{>0} to mathbb{R}^{>0}$, with the restriction $x,y > 0$ in the functional equation which is not the original problem. Actually it is probably harder because $y > 0$ is an annoying constraint.



    Because $f$ is positive valued, we know that
    $$ f(x^2 + y) > f(x)^2 tag{1} $$
    and
    $$f(x^2 + y) > frac{f(xy)}{f(x)} tag{2}. $$
    The following two Observations motivate us to compute $f(1)$ somewhat "analytically":




    1. Taking $y = 1$ gives $f(x^2 + 1) > 1$, i.e. $f(x) > 1$ for $x > 1$.

    2. The solutions to $x^2 + y = x$ are of the form $(x,x - x^2)$ where $x < 1$. Then from (1) we conclude that for any $x < 1$, $f(x) > f(x)^2$ i.e. $f(x) < 1$.


    Let's repeat the latter trick, plugging into the actual equation. Then for $x < 1$ we have
    $$f(x) = f(x)^2 + frac{f(x^2 - x^3)}{f(x)}$$
    and rearranging,
    $$ f(x^2 - x^3) = f(x)^2 - f(x)^3. $$



    Let $u_0 = 1/2$ (or an arbitrary value below $1$) and inductively let $u_{i + 1} = u_i^2 - u_i^3$. Because $u_{i + 1} < u_i^2 < 1$ and $u_{i + 1} = u_i^2 - u_i^3 > 0$ we have that $lim_{i to infty} u_i = 0$, and a similar argument gives that $lim_{i to infty} f(u_i) = 0$. Now
    $$ f(1 + u_i) = f(1)^2 + frac{f(u_i)}{f(1)} $$
    and taking the limit
    $$ lim_{i to infty} f(1 + u_i) = f(1)^2. $$
    Because $1 < f(1 + u_i)$ by Observation 1 we get that $1 le f(1)^2$ hence $1 le f(1)$.



    Now go back to the equality and plug in $y = 1$. We get
    $$f(x^2 + 1) = f(x)^2 + 1$$
    and taking $x = u_i$ and a limit
    $$lim_{i to infty} f(u_i^2 + 1) = 1. $$
    Taking $x = 1$, $y = u_i^2$ in the equality
    $$ f(1 + u_i^2) = f(1)^2 + frac{f(u_i^2)}{f(1)} $$
    and taking the limit we get
    $$ 1 = lim_{i to infty} left(f(1)^2 + frac{f(u_i^2)}{f(1)}right) $$
    so in particular $f(1)^2 le 1$, i.e. $f(1) le 1$.



    So finally we have shown $f(1) = 1$. Taking $x = 1$
    $$ f(1 + y) = f(1)^2 + frac{f(y)}{f(1)} = 1 + f(y) tag{3}$$
    so by induction
    $$f(n + r) = n + f(r)$$
    for $n in mathbb{Z}^{> 0}$, $r ge 0$.
    Taking $y = 1$ in the original equation
    $$ f(x^2 + 1) = f(x)^2 + 1 $$
    so by (3), for all $x$ $f(x^2) = f(x)^2$, i.e. $f(sqrt{x}) = sqrt{f(x)}$. Now recalling the original equation
    $$ f(x^2 + y) = f(x)^2 + frac{f(xy)}{f(x)} = f(x^2) + frac{f(xy)}{f(x)} tag{4} $$
    so $f$ is an increasing function. Recalling our sequence $u_i$ with $u_i to 0, f(u_i) to 0$ as $i to infty$, we conclude that $lim_{x to^+ 0} f(x) = 0$. Taking the limit in (4),
    $$ lim_{y to^+ 0} f(x^2 + y) = f(x^2) $$
    and so $f$ is right-continuous everywhere.
    Fix arbitrary $z$, and let $x^2 = z - y$.
    Then
    $$ f(z) - f(z - y) = f(x^2 + y) - f(x^2) = frac{f(xy)}{f(x)} < frac{f(sqrt{z} cdot y)}{f(sqrt{z - k})} $$
    for any fixed $k > 0$ such that $0 < z - k$, for all $y < k$ (so $sqrt{z - k} < sqrt{z - y}$). Taking the limit as $y to^+ 0$ we find
    that $f$ is left-continuous at $z$. Hence $f$ is continuous everywhere.



    I now use the (nontrivial, but intuitive) fact that non-negative multiples of $sqrt{2}$ are dense in $mathbb{R}/mathbb{Z}$. In other words
    $$ {asqrt{2} + b : a in mathbb{Z}^{ge 0},b in mathbb{Z}} $$
    is dense in $mathbb{R}$. For any positive $asqrt{2} + b = sqrt{2a^2} + b$, if $b ge 0$ then
    we know
    $$ f(sqrt{2a^2} + b) = b + f(sqrt{2a^2}) = b + sqrt{2a^2}. $$
    If $b < 0$ then we know
    $$ f(sqrt{2a^2} + b) - b = f(sqrt{2a^2} + b - b) = sqrt{2a^2}. $$
    So $f$ equals the identity on a dense subset of $mathbb{R}^{> 0}$. By continuity, $f = id$.



    PS This might not be the shortest solution! At least it is fairly systematic. Also, my apologies, there may be small mistakes, so caveat emptor.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Apr 13 '17 at 12:19









    Community

    1




    1










    answered Jun 14 '14 at 0:07









    Frederic KoehlerFrederic Koehler

    1066




    1066












    • This is a very nice and titanic solution :P but the first assumption, namely that f is positively valued, isn't an hypothesis in the original formulation from what I understand. Anyway there are a couple of very nice methods here that I'm sure are fruitful and will keep me busy revisiting and adapting them. PD: I can't upvote your response because I don't have enough reputation. Thank you VERY much for your time bro.
      – user430825
      Jun 14 '14 at 1:57












    • Oops! I misread the problem. At least you know how to solve the problem when f is positive-valued now. I will think about it when f is on R^*, possibly there is also a shorter solution then, haha.
      – Frederic Koehler
      Jun 14 '14 at 2:10










    • @user430825 I have uploaded a (hopefully correct) solution to the correct problem, but feel free to try to solve it on your own first! the problem was fun, you're welcome.
      – Frederic Koehler
      Jun 14 '14 at 4:54










    • very clever my friend, the part where you show that it cannot be an even function, I was working on that one! hehe. I'll try to come up with a solution that doesn't appeal to your special case! Thanks again for your time.
      – user430825
      Jun 14 '14 at 10:58


















    • This is a very nice and titanic solution :P but the first assumption, namely that f is positively valued, isn't an hypothesis in the original formulation from what I understand. Anyway there are a couple of very nice methods here that I'm sure are fruitful and will keep me busy revisiting and adapting them. PD: I can't upvote your response because I don't have enough reputation. Thank you VERY much for your time bro.
      – user430825
      Jun 14 '14 at 1:57












    • Oops! I misread the problem. At least you know how to solve the problem when f is positive-valued now. I will think about it when f is on R^*, possibly there is also a shorter solution then, haha.
      – Frederic Koehler
      Jun 14 '14 at 2:10










    • @user430825 I have uploaded a (hopefully correct) solution to the correct problem, but feel free to try to solve it on your own first! the problem was fun, you're welcome.
      – Frederic Koehler
      Jun 14 '14 at 4:54










    • very clever my friend, the part where you show that it cannot be an even function, I was working on that one! hehe. I'll try to come up with a solution that doesn't appeal to your special case! Thanks again for your time.
      – user430825
      Jun 14 '14 at 10:58
















    This is a very nice and titanic solution :P but the first assumption, namely that f is positively valued, isn't an hypothesis in the original formulation from what I understand. Anyway there are a couple of very nice methods here that I'm sure are fruitful and will keep me busy revisiting and adapting them. PD: I can't upvote your response because I don't have enough reputation. Thank you VERY much for your time bro.
    – user430825
    Jun 14 '14 at 1:57






    This is a very nice and titanic solution :P but the first assumption, namely that f is positively valued, isn't an hypothesis in the original formulation from what I understand. Anyway there are a couple of very nice methods here that I'm sure are fruitful and will keep me busy revisiting and adapting them. PD: I can't upvote your response because I don't have enough reputation. Thank you VERY much for your time bro.
    – user430825
    Jun 14 '14 at 1:57














    Oops! I misread the problem. At least you know how to solve the problem when f is positive-valued now. I will think about it when f is on R^*, possibly there is also a shorter solution then, haha.
    – Frederic Koehler
    Jun 14 '14 at 2:10




    Oops! I misread the problem. At least you know how to solve the problem when f is positive-valued now. I will think about it when f is on R^*, possibly there is also a shorter solution then, haha.
    – Frederic Koehler
    Jun 14 '14 at 2:10












    @user430825 I have uploaded a (hopefully correct) solution to the correct problem, but feel free to try to solve it on your own first! the problem was fun, you're welcome.
    – Frederic Koehler
    Jun 14 '14 at 4:54




    @user430825 I have uploaded a (hopefully correct) solution to the correct problem, but feel free to try to solve it on your own first! the problem was fun, you're welcome.
    – Frederic Koehler
    Jun 14 '14 at 4:54












    very clever my friend, the part where you show that it cannot be an even function, I was working on that one! hehe. I'll try to come up with a solution that doesn't appeal to your special case! Thanks again for your time.
    – user430825
    Jun 14 '14 at 10:58




    very clever my friend, the part where you show that it cannot be an even function, I was working on that one! hehe. I'll try to come up with a solution that doesn't appeal to your special case! Thanks again for your time.
    – user430825
    Jun 14 '14 at 10:58











    4














    This might not be the optimal approach... but it attempts to avoid reasoning based on real analysis (well, apart from the very definition of irrational numbers).



    Determine the value of $f(1)$



    Let $f(1)=A$. Then, for any $y$ different from $0$ and $(-1)$ we have $$f(1+y)=f(1)^2 + f(y)/f(1) = A^2 + f(y)/A$$ which allows us to express the following quantities: $$begin{eqnarray}
    f(2) & = & A^2 + 1 \
    f(3) & = & A^2 + A + frac{1}{A} \
    f(4) & = & A^2 + A + 1 + frac{1}{A^2} \
    f(5) & = & A^2 + A + 1 + frac{1}{A} + frac{1}{A^3} \
    end{eqnarray}$$



    At the same time, we have $f(x^2+1)=f(x)^2+f(x)/f(x)=f(x)^2+1$. This gives us $$f(5)=f(2)^2+1$$ Equating the two expressions for $f(5)$ and substituting for $f(2)$ gives us
    $$(A^2+1)^2 + 1 = A^2+A+1+frac{1}{A} + frac{1}{A^3}$$ which simplifies to $$begin{eqnarray}0 & = & A^7 + A^5 - A^4 + A^3 - A^2 - 1 \ & = & (A-1)(A^2-A+1)(A^2+A+1)^2end{eqnarray}$$ whose only real root is $A=1$. Thus, $f(1)=1$ and we have $f(1+y)=1+f(y)$ for all $y$ other than $0$ and $(-1)$. In particular, this means that $f(k)=k$ for any positive integer $k$.



    Show additive property for numbers of same sign



    Now, consider real numbers $u,v$ of the same sign and these two equalities:
    $$begin{eqnarray}
    f(v^2 + u/v) & = & f(v)^2 + f(u)/f(v) \
    f(v^2 + (u/v + 1)) & = & f(v)^2 + f(u+v)/f(v) \
    end{eqnarray}$$
    We can apply the equality $f(1+y)=1+f(y)$ since both $v^2$ and $u/v$ are positive to obtain
    $$f(u+v)=f(u)+f(v)$$



    Note: This equation is very similar to Cauchy's functional equation; a quite interesting problem of its own. The equality given in our problem is stronger than "just" plain additivity, though.



    Determine values of $f$ on positive rationals



    It's quite easy to see that this equality implies that $fleft(frac{p}{q}right)=frac{p}{q}$ for any positive integers $p,q$; one simply expands $$p=f(p)=fleft(frac{p}{q}right)+fleft(frac{p}{q}right)+ldots+fleft(frac{p}{q}right)=qcdot fleft(frac{p}{q}right)$$



    Show that $f$ is increasing on positive reals



    For $y>0$, we can write $$begin{eqnarray}
    f(y) = f(y)+1-1 & = & f(y+1)-1 \
    & = & fleft((sqrt{y})^2+1right)-1 = \
    & = & fleft(sqrt{y}right)^2+1-1 \
    & = & fleft(sqrt{y}right)^2>0
    end{eqnarray}$$



    Thus, for any positive reals $u>v>0$, we get $$f(u)=f(v+(u-v))=f(v)+f(u-v)>f(v)$$



    Determine the values of $f$ for positive reals



    Let's show that $f(x)=x$ for $x>0$. If that wasn't the case, we'd have either $f(x)>x$ or $0<f(x)<x$. In the first case, we could find rational number $z$ with $f(x)>z>x$. First inequality implies $f(x)>f(z)$ since $f(z)=z$, while the second one implies $f(z)>f(x)$, a contradiction. The other case can be treated in the same way. Thus, $f(x)=x$.



    Alternate approach is to realize that the Dedekind cut corresponding to irrational $x>0$ must be the same as the Dedekind cut corresponding to $f(x)$.



    Finish the proof by considering the negative reals



    If $x<0$ is non-integer, we can simply use the equality $f(1+y)=1+f(y)$ to show that $f(x)=x$; just by adding $1$ until we reach a positive number. For negative integers, we can bootstrap the process by considering $f(-1)=fleft(-frac{1}{2}right)+fleft(-frac{1}{2}right)=(-1)$ and do the same by reaching $(-1)$.



    The only function satisfying the given condition is $f(x)=x$ Q.E.D.






    share|cite|improve this answer





















    • Excellent answer. Really nice. Thanks for your time!
      – user430825
      Jun 14 '14 at 17:22


















    4














    This might not be the optimal approach... but it attempts to avoid reasoning based on real analysis (well, apart from the very definition of irrational numbers).



    Determine the value of $f(1)$



    Let $f(1)=A$. Then, for any $y$ different from $0$ and $(-1)$ we have $$f(1+y)=f(1)^2 + f(y)/f(1) = A^2 + f(y)/A$$ which allows us to express the following quantities: $$begin{eqnarray}
    f(2) & = & A^2 + 1 \
    f(3) & = & A^2 + A + frac{1}{A} \
    f(4) & = & A^2 + A + 1 + frac{1}{A^2} \
    f(5) & = & A^2 + A + 1 + frac{1}{A} + frac{1}{A^3} \
    end{eqnarray}$$



    At the same time, we have $f(x^2+1)=f(x)^2+f(x)/f(x)=f(x)^2+1$. This gives us $$f(5)=f(2)^2+1$$ Equating the two expressions for $f(5)$ and substituting for $f(2)$ gives us
    $$(A^2+1)^2 + 1 = A^2+A+1+frac{1}{A} + frac{1}{A^3}$$ which simplifies to $$begin{eqnarray}0 & = & A^7 + A^5 - A^4 + A^3 - A^2 - 1 \ & = & (A-1)(A^2-A+1)(A^2+A+1)^2end{eqnarray}$$ whose only real root is $A=1$. Thus, $f(1)=1$ and we have $f(1+y)=1+f(y)$ for all $y$ other than $0$ and $(-1)$. In particular, this means that $f(k)=k$ for any positive integer $k$.



    Show additive property for numbers of same sign



    Now, consider real numbers $u,v$ of the same sign and these two equalities:
    $$begin{eqnarray}
    f(v^2 + u/v) & = & f(v)^2 + f(u)/f(v) \
    f(v^2 + (u/v + 1)) & = & f(v)^2 + f(u+v)/f(v) \
    end{eqnarray}$$
    We can apply the equality $f(1+y)=1+f(y)$ since both $v^2$ and $u/v$ are positive to obtain
    $$f(u+v)=f(u)+f(v)$$



    Note: This equation is very similar to Cauchy's functional equation; a quite interesting problem of its own. The equality given in our problem is stronger than "just" plain additivity, though.



    Determine values of $f$ on positive rationals



    It's quite easy to see that this equality implies that $fleft(frac{p}{q}right)=frac{p}{q}$ for any positive integers $p,q$; one simply expands $$p=f(p)=fleft(frac{p}{q}right)+fleft(frac{p}{q}right)+ldots+fleft(frac{p}{q}right)=qcdot fleft(frac{p}{q}right)$$



    Show that $f$ is increasing on positive reals



    For $y>0$, we can write $$begin{eqnarray}
    f(y) = f(y)+1-1 & = & f(y+1)-1 \
    & = & fleft((sqrt{y})^2+1right)-1 = \
    & = & fleft(sqrt{y}right)^2+1-1 \
    & = & fleft(sqrt{y}right)^2>0
    end{eqnarray}$$



    Thus, for any positive reals $u>v>0$, we get $$f(u)=f(v+(u-v))=f(v)+f(u-v)>f(v)$$



    Determine the values of $f$ for positive reals



    Let's show that $f(x)=x$ for $x>0$. If that wasn't the case, we'd have either $f(x)>x$ or $0<f(x)<x$. In the first case, we could find rational number $z$ with $f(x)>z>x$. First inequality implies $f(x)>f(z)$ since $f(z)=z$, while the second one implies $f(z)>f(x)$, a contradiction. The other case can be treated in the same way. Thus, $f(x)=x$.



    Alternate approach is to realize that the Dedekind cut corresponding to irrational $x>0$ must be the same as the Dedekind cut corresponding to $f(x)$.



    Finish the proof by considering the negative reals



    If $x<0$ is non-integer, we can simply use the equality $f(1+y)=1+f(y)$ to show that $f(x)=x$; just by adding $1$ until we reach a positive number. For negative integers, we can bootstrap the process by considering $f(-1)=fleft(-frac{1}{2}right)+fleft(-frac{1}{2}right)=(-1)$ and do the same by reaching $(-1)$.



    The only function satisfying the given condition is $f(x)=x$ Q.E.D.






    share|cite|improve this answer





















    • Excellent answer. Really nice. Thanks for your time!
      – user430825
      Jun 14 '14 at 17:22
















    4












    4








    4






    This might not be the optimal approach... but it attempts to avoid reasoning based on real analysis (well, apart from the very definition of irrational numbers).



    Determine the value of $f(1)$



    Let $f(1)=A$. Then, for any $y$ different from $0$ and $(-1)$ we have $$f(1+y)=f(1)^2 + f(y)/f(1) = A^2 + f(y)/A$$ which allows us to express the following quantities: $$begin{eqnarray}
    f(2) & = & A^2 + 1 \
    f(3) & = & A^2 + A + frac{1}{A} \
    f(4) & = & A^2 + A + 1 + frac{1}{A^2} \
    f(5) & = & A^2 + A + 1 + frac{1}{A} + frac{1}{A^3} \
    end{eqnarray}$$



    At the same time, we have $f(x^2+1)=f(x)^2+f(x)/f(x)=f(x)^2+1$. This gives us $$f(5)=f(2)^2+1$$ Equating the two expressions for $f(5)$ and substituting for $f(2)$ gives us
    $$(A^2+1)^2 + 1 = A^2+A+1+frac{1}{A} + frac{1}{A^3}$$ which simplifies to $$begin{eqnarray}0 & = & A^7 + A^5 - A^4 + A^3 - A^2 - 1 \ & = & (A-1)(A^2-A+1)(A^2+A+1)^2end{eqnarray}$$ whose only real root is $A=1$. Thus, $f(1)=1$ and we have $f(1+y)=1+f(y)$ for all $y$ other than $0$ and $(-1)$. In particular, this means that $f(k)=k$ for any positive integer $k$.



    Show additive property for numbers of same sign



    Now, consider real numbers $u,v$ of the same sign and these two equalities:
    $$begin{eqnarray}
    f(v^2 + u/v) & = & f(v)^2 + f(u)/f(v) \
    f(v^2 + (u/v + 1)) & = & f(v)^2 + f(u+v)/f(v) \
    end{eqnarray}$$
    We can apply the equality $f(1+y)=1+f(y)$ since both $v^2$ and $u/v$ are positive to obtain
    $$f(u+v)=f(u)+f(v)$$



    Note: This equation is very similar to Cauchy's functional equation; a quite interesting problem of its own. The equality given in our problem is stronger than "just" plain additivity, though.



    Determine values of $f$ on positive rationals



    It's quite easy to see that this equality implies that $fleft(frac{p}{q}right)=frac{p}{q}$ for any positive integers $p,q$; one simply expands $$p=f(p)=fleft(frac{p}{q}right)+fleft(frac{p}{q}right)+ldots+fleft(frac{p}{q}right)=qcdot fleft(frac{p}{q}right)$$



    Show that $f$ is increasing on positive reals



    For $y>0$, we can write $$begin{eqnarray}
    f(y) = f(y)+1-1 & = & f(y+1)-1 \
    & = & fleft((sqrt{y})^2+1right)-1 = \
    & = & fleft(sqrt{y}right)^2+1-1 \
    & = & fleft(sqrt{y}right)^2>0
    end{eqnarray}$$



    Thus, for any positive reals $u>v>0$, we get $$f(u)=f(v+(u-v))=f(v)+f(u-v)>f(v)$$



    Determine the values of $f$ for positive reals



    Let's show that $f(x)=x$ for $x>0$. If that wasn't the case, we'd have either $f(x)>x$ or $0<f(x)<x$. In the first case, we could find rational number $z$ with $f(x)>z>x$. First inequality implies $f(x)>f(z)$ since $f(z)=z$, while the second one implies $f(z)>f(x)$, a contradiction. The other case can be treated in the same way. Thus, $f(x)=x$.



    Alternate approach is to realize that the Dedekind cut corresponding to irrational $x>0$ must be the same as the Dedekind cut corresponding to $f(x)$.



    Finish the proof by considering the negative reals



    If $x<0$ is non-integer, we can simply use the equality $f(1+y)=1+f(y)$ to show that $f(x)=x$; just by adding $1$ until we reach a positive number. For negative integers, we can bootstrap the process by considering $f(-1)=fleft(-frac{1}{2}right)+fleft(-frac{1}{2}right)=(-1)$ and do the same by reaching $(-1)$.



    The only function satisfying the given condition is $f(x)=x$ Q.E.D.






    share|cite|improve this answer












    This might not be the optimal approach... but it attempts to avoid reasoning based on real analysis (well, apart from the very definition of irrational numbers).



    Determine the value of $f(1)$



    Let $f(1)=A$. Then, for any $y$ different from $0$ and $(-1)$ we have $$f(1+y)=f(1)^2 + f(y)/f(1) = A^2 + f(y)/A$$ which allows us to express the following quantities: $$begin{eqnarray}
    f(2) & = & A^2 + 1 \
    f(3) & = & A^2 + A + frac{1}{A} \
    f(4) & = & A^2 + A + 1 + frac{1}{A^2} \
    f(5) & = & A^2 + A + 1 + frac{1}{A} + frac{1}{A^3} \
    end{eqnarray}$$



    At the same time, we have $f(x^2+1)=f(x)^2+f(x)/f(x)=f(x)^2+1$. This gives us $$f(5)=f(2)^2+1$$ Equating the two expressions for $f(5)$ and substituting for $f(2)$ gives us
    $$(A^2+1)^2 + 1 = A^2+A+1+frac{1}{A} + frac{1}{A^3}$$ which simplifies to $$begin{eqnarray}0 & = & A^7 + A^5 - A^4 + A^3 - A^2 - 1 \ & = & (A-1)(A^2-A+1)(A^2+A+1)^2end{eqnarray}$$ whose only real root is $A=1$. Thus, $f(1)=1$ and we have $f(1+y)=1+f(y)$ for all $y$ other than $0$ and $(-1)$. In particular, this means that $f(k)=k$ for any positive integer $k$.



    Show additive property for numbers of same sign



    Now, consider real numbers $u,v$ of the same sign and these two equalities:
    $$begin{eqnarray}
    f(v^2 + u/v) & = & f(v)^2 + f(u)/f(v) \
    f(v^2 + (u/v + 1)) & = & f(v)^2 + f(u+v)/f(v) \
    end{eqnarray}$$
    We can apply the equality $f(1+y)=1+f(y)$ since both $v^2$ and $u/v$ are positive to obtain
    $$f(u+v)=f(u)+f(v)$$



    Note: This equation is very similar to Cauchy's functional equation; a quite interesting problem of its own. The equality given in our problem is stronger than "just" plain additivity, though.



    Determine values of $f$ on positive rationals



    It's quite easy to see that this equality implies that $fleft(frac{p}{q}right)=frac{p}{q}$ for any positive integers $p,q$; one simply expands $$p=f(p)=fleft(frac{p}{q}right)+fleft(frac{p}{q}right)+ldots+fleft(frac{p}{q}right)=qcdot fleft(frac{p}{q}right)$$



    Show that $f$ is increasing on positive reals



    For $y>0$, we can write $$begin{eqnarray}
    f(y) = f(y)+1-1 & = & f(y+1)-1 \
    & = & fleft((sqrt{y})^2+1right)-1 = \
    & = & fleft(sqrt{y}right)^2+1-1 \
    & = & fleft(sqrt{y}right)^2>0
    end{eqnarray}$$



    Thus, for any positive reals $u>v>0$, we get $$f(u)=f(v+(u-v))=f(v)+f(u-v)>f(v)$$



    Determine the values of $f$ for positive reals



    Let's show that $f(x)=x$ for $x>0$. If that wasn't the case, we'd have either $f(x)>x$ or $0<f(x)<x$. In the first case, we could find rational number $z$ with $f(x)>z>x$. First inequality implies $f(x)>f(z)$ since $f(z)=z$, while the second one implies $f(z)>f(x)$, a contradiction. The other case can be treated in the same way. Thus, $f(x)=x$.



    Alternate approach is to realize that the Dedekind cut corresponding to irrational $x>0$ must be the same as the Dedekind cut corresponding to $f(x)$.



    Finish the proof by considering the negative reals



    If $x<0$ is non-integer, we can simply use the equality $f(1+y)=1+f(y)$ to show that $f(x)=x$; just by adding $1$ until we reach a positive number. For negative integers, we can bootstrap the process by considering $f(-1)=fleft(-frac{1}{2}right)+fleft(-frac{1}{2}right)=(-1)$ and do the same by reaching $(-1)$.



    The only function satisfying the given condition is $f(x)=x$ Q.E.D.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jun 14 '14 at 16:49









    Peter KošinárPeter Košinár

    8,13111434




    8,13111434












    • Excellent answer. Really nice. Thanks for your time!
      – user430825
      Jun 14 '14 at 17:22




















    • Excellent answer. Really nice. Thanks for your time!
      – user430825
      Jun 14 '14 at 17:22


















    Excellent answer. Really nice. Thanks for your time!
    – user430825
    Jun 14 '14 at 17:22






    Excellent answer. Really nice. Thanks for your time!
    – user430825
    Jun 14 '14 at 17:22













    0














    $newcommand{NN}{{mathbb{N}}}newcommand{CC}{{mathbb{C}}}newcommand{RR}{{mathbb{R}}}newcommand{QQ}{{mathbb Q}}newcommand{ra}{rightarrow}newcommand{ds}{displaystyle}newcommand{mat}[1]{left(begin{matrix}#1end{matrix}right)}newcommand{lam}{lambda}renewcommand{P}{{mathcal P}}newcommand{N}{{mathcal N}}newcommand{M}{{mathcal M}}renewcommand{L}{{mathcal L}}newcommand{EQ}[2]{begin{equation} {{#2}label{#1}} end{equation}}newcommand{eps}{varepsilon}$
    It contains numerous elements that are also in other answers - I put it anyway.
    Observe that $f$ cannot be constant, because the equation $C=C^2+1$ has no real solution.



    Putting $y=1$, we find that $f(x)^2+1=f(x^2+1)=f(-x)^2+1$ and hence $f(x)^2=f(-x)^2$
    for all $xin A$. Putting $x=pm1$, we obtain $f(-y)/f(-1)=f(y)/f(1)$ for all $y$. So either
    $f(-x)=f(x)$ for all $x$ or $f(-x)=-f(x)$ for all $x$.



    The solution $y$ of $x^2+y=xy$ is $y=frac{x^2}{x-1}$ unless $x=1$. Using this $y$ we have
    $$fleft(frac{x^3}{x-1}right)=frac{f(x)^3}{f(x)-1}mbox{ for }xneq1.$$
    Consider now a solution $z$ of $frac{z^2}{z-1}=-1$. Then $f(-z)=frac{f(z)^3}{f(z)-1}$.
    As a consequence $f(-z)$ cannot equal $f(z)$ because $1=frac{f(z)^2}{f(z)-1}$ has no real
    solution for $f(z)$. Hence $f(-z)=-f(z)$ and, as seen above, $f(-x)=-f(x)$ for all $x.$



    Next consider $x=1$ and replace $y$ by $-y$. We find $f(1-y)=f(1)^2-frac{f(y)}{f(1)}$
    for $yneq1$. Replacing here $y$ by $1-y$ yields with $w=f(1)$
    $$w^2f(y)=w^4-wf(1-y)=w^4-w^3+f(y)mbox{ for all }yin A,yneq1.$$
    Thus $(w^2-1)f(y)=w^4-w^3$. So we must have $w=1$ as otherwise $w=-1$
    which leads to a contradiction or $f(y)=frac{w^3}{w+1}$ is constant which had been excluded.
    So finally we proved that $w=f(1)=1$.



    Putting $x=1$ now gives $f(y+1)=f(y)+1$ for $yneq-1$. In particular, we find $f(n)=n$ and
    $f(-n)=-n$ for all positive integers $n$. Putting $x=n$, we obtain
    $f(y+n^2)=n^2+frac1nf(ny)$ for all $yneq-n^2$ and $f(ny)=nf(y)$ for all positive $y$
    (In fact, we only have to exclude that $y$ is a negative integer).
    In particular this leads to $f(q)=q$ for all rational positive $q$. It is extended to
    negative rationals using $f(-x)=-f(x)$.
    Moreover, we obtain $f(y+m)=f(y)+m$ for all non-integer $y$ and thus
    $f(y+frac mn)=frac1nf(ny+m)=frac1n(nf(y)+m)=f(y)+frac mn$
    for all irrational $y$ and hence $f(y+q)=f(y)+q$
    for all rational $q$ and irrational $y$.



    Finally, as $f(x^2+1)=f(x)^2+1$ and hence $f(x^2)=f(x)^2$ for all $x$,
    the values of $f$ for positive $y$ are positive and hence also its values for
    negative $y$ are negative.
    Given any irrational $y$, we consider rational $q,r$ with $q<y<r$. Then
    as $y-q>0$, we have $f(y)-q=f(y-q)>0$ and hence $q<f(y)$. In the same way we show that
    $f(y)<r$. Since we can use any rational $q,r$ with $q<y<r$,
    we have shown that $f(y)=y$ also for irrational $y$.






    share|cite|improve this answer




























      0














      $newcommand{NN}{{mathbb{N}}}newcommand{CC}{{mathbb{C}}}newcommand{RR}{{mathbb{R}}}newcommand{QQ}{{mathbb Q}}newcommand{ra}{rightarrow}newcommand{ds}{displaystyle}newcommand{mat}[1]{left(begin{matrix}#1end{matrix}right)}newcommand{lam}{lambda}renewcommand{P}{{mathcal P}}newcommand{N}{{mathcal N}}newcommand{M}{{mathcal M}}renewcommand{L}{{mathcal L}}newcommand{EQ}[2]{begin{equation} {{#2}label{#1}} end{equation}}newcommand{eps}{varepsilon}$
      It contains numerous elements that are also in other answers - I put it anyway.
      Observe that $f$ cannot be constant, because the equation $C=C^2+1$ has no real solution.



      Putting $y=1$, we find that $f(x)^2+1=f(x^2+1)=f(-x)^2+1$ and hence $f(x)^2=f(-x)^2$
      for all $xin A$. Putting $x=pm1$, we obtain $f(-y)/f(-1)=f(y)/f(1)$ for all $y$. So either
      $f(-x)=f(x)$ for all $x$ or $f(-x)=-f(x)$ for all $x$.



      The solution $y$ of $x^2+y=xy$ is $y=frac{x^2}{x-1}$ unless $x=1$. Using this $y$ we have
      $$fleft(frac{x^3}{x-1}right)=frac{f(x)^3}{f(x)-1}mbox{ for }xneq1.$$
      Consider now a solution $z$ of $frac{z^2}{z-1}=-1$. Then $f(-z)=frac{f(z)^3}{f(z)-1}$.
      As a consequence $f(-z)$ cannot equal $f(z)$ because $1=frac{f(z)^2}{f(z)-1}$ has no real
      solution for $f(z)$. Hence $f(-z)=-f(z)$ and, as seen above, $f(-x)=-f(x)$ for all $x.$



      Next consider $x=1$ and replace $y$ by $-y$. We find $f(1-y)=f(1)^2-frac{f(y)}{f(1)}$
      for $yneq1$. Replacing here $y$ by $1-y$ yields with $w=f(1)$
      $$w^2f(y)=w^4-wf(1-y)=w^4-w^3+f(y)mbox{ for all }yin A,yneq1.$$
      Thus $(w^2-1)f(y)=w^4-w^3$. So we must have $w=1$ as otherwise $w=-1$
      which leads to a contradiction or $f(y)=frac{w^3}{w+1}$ is constant which had been excluded.
      So finally we proved that $w=f(1)=1$.



      Putting $x=1$ now gives $f(y+1)=f(y)+1$ for $yneq-1$. In particular, we find $f(n)=n$ and
      $f(-n)=-n$ for all positive integers $n$. Putting $x=n$, we obtain
      $f(y+n^2)=n^2+frac1nf(ny)$ for all $yneq-n^2$ and $f(ny)=nf(y)$ for all positive $y$
      (In fact, we only have to exclude that $y$ is a negative integer).
      In particular this leads to $f(q)=q$ for all rational positive $q$. It is extended to
      negative rationals using $f(-x)=-f(x)$.
      Moreover, we obtain $f(y+m)=f(y)+m$ for all non-integer $y$ and thus
      $f(y+frac mn)=frac1nf(ny+m)=frac1n(nf(y)+m)=f(y)+frac mn$
      for all irrational $y$ and hence $f(y+q)=f(y)+q$
      for all rational $q$ and irrational $y$.



      Finally, as $f(x^2+1)=f(x)^2+1$ and hence $f(x^2)=f(x)^2$ for all $x$,
      the values of $f$ for positive $y$ are positive and hence also its values for
      negative $y$ are negative.
      Given any irrational $y$, we consider rational $q,r$ with $q<y<r$. Then
      as $y-q>0$, we have $f(y)-q=f(y-q)>0$ and hence $q<f(y)$. In the same way we show that
      $f(y)<r$. Since we can use any rational $q,r$ with $q<y<r$,
      we have shown that $f(y)=y$ also for irrational $y$.






      share|cite|improve this answer


























        0












        0








        0






        $newcommand{NN}{{mathbb{N}}}newcommand{CC}{{mathbb{C}}}newcommand{RR}{{mathbb{R}}}newcommand{QQ}{{mathbb Q}}newcommand{ra}{rightarrow}newcommand{ds}{displaystyle}newcommand{mat}[1]{left(begin{matrix}#1end{matrix}right)}newcommand{lam}{lambda}renewcommand{P}{{mathcal P}}newcommand{N}{{mathcal N}}newcommand{M}{{mathcal M}}renewcommand{L}{{mathcal L}}newcommand{EQ}[2]{begin{equation} {{#2}label{#1}} end{equation}}newcommand{eps}{varepsilon}$
        It contains numerous elements that are also in other answers - I put it anyway.
        Observe that $f$ cannot be constant, because the equation $C=C^2+1$ has no real solution.



        Putting $y=1$, we find that $f(x)^2+1=f(x^2+1)=f(-x)^2+1$ and hence $f(x)^2=f(-x)^2$
        for all $xin A$. Putting $x=pm1$, we obtain $f(-y)/f(-1)=f(y)/f(1)$ for all $y$. So either
        $f(-x)=f(x)$ for all $x$ or $f(-x)=-f(x)$ for all $x$.



        The solution $y$ of $x^2+y=xy$ is $y=frac{x^2}{x-1}$ unless $x=1$. Using this $y$ we have
        $$fleft(frac{x^3}{x-1}right)=frac{f(x)^3}{f(x)-1}mbox{ for }xneq1.$$
        Consider now a solution $z$ of $frac{z^2}{z-1}=-1$. Then $f(-z)=frac{f(z)^3}{f(z)-1}$.
        As a consequence $f(-z)$ cannot equal $f(z)$ because $1=frac{f(z)^2}{f(z)-1}$ has no real
        solution for $f(z)$. Hence $f(-z)=-f(z)$ and, as seen above, $f(-x)=-f(x)$ for all $x.$



        Next consider $x=1$ and replace $y$ by $-y$. We find $f(1-y)=f(1)^2-frac{f(y)}{f(1)}$
        for $yneq1$. Replacing here $y$ by $1-y$ yields with $w=f(1)$
        $$w^2f(y)=w^4-wf(1-y)=w^4-w^3+f(y)mbox{ for all }yin A,yneq1.$$
        Thus $(w^2-1)f(y)=w^4-w^3$. So we must have $w=1$ as otherwise $w=-1$
        which leads to a contradiction or $f(y)=frac{w^3}{w+1}$ is constant which had been excluded.
        So finally we proved that $w=f(1)=1$.



        Putting $x=1$ now gives $f(y+1)=f(y)+1$ for $yneq-1$. In particular, we find $f(n)=n$ and
        $f(-n)=-n$ for all positive integers $n$. Putting $x=n$, we obtain
        $f(y+n^2)=n^2+frac1nf(ny)$ for all $yneq-n^2$ and $f(ny)=nf(y)$ for all positive $y$
        (In fact, we only have to exclude that $y$ is a negative integer).
        In particular this leads to $f(q)=q$ for all rational positive $q$. It is extended to
        negative rationals using $f(-x)=-f(x)$.
        Moreover, we obtain $f(y+m)=f(y)+m$ for all non-integer $y$ and thus
        $f(y+frac mn)=frac1nf(ny+m)=frac1n(nf(y)+m)=f(y)+frac mn$
        for all irrational $y$ and hence $f(y+q)=f(y)+q$
        for all rational $q$ and irrational $y$.



        Finally, as $f(x^2+1)=f(x)^2+1$ and hence $f(x^2)=f(x)^2$ for all $x$,
        the values of $f$ for positive $y$ are positive and hence also its values for
        negative $y$ are negative.
        Given any irrational $y$, we consider rational $q,r$ with $q<y<r$. Then
        as $y-q>0$, we have $f(y)-q=f(y-q)>0$ and hence $q<f(y)$. In the same way we show that
        $f(y)<r$. Since we can use any rational $q,r$ with $q<y<r$,
        we have shown that $f(y)=y$ also for irrational $y$.






        share|cite|improve this answer














        $newcommand{NN}{{mathbb{N}}}newcommand{CC}{{mathbb{C}}}newcommand{RR}{{mathbb{R}}}newcommand{QQ}{{mathbb Q}}newcommand{ra}{rightarrow}newcommand{ds}{displaystyle}newcommand{mat}[1]{left(begin{matrix}#1end{matrix}right)}newcommand{lam}{lambda}renewcommand{P}{{mathcal P}}newcommand{N}{{mathcal N}}newcommand{M}{{mathcal M}}renewcommand{L}{{mathcal L}}newcommand{EQ}[2]{begin{equation} {{#2}label{#1}} end{equation}}newcommand{eps}{varepsilon}$
        It contains numerous elements that are also in other answers - I put it anyway.
        Observe that $f$ cannot be constant, because the equation $C=C^2+1$ has no real solution.



        Putting $y=1$, we find that $f(x)^2+1=f(x^2+1)=f(-x)^2+1$ and hence $f(x)^2=f(-x)^2$
        for all $xin A$. Putting $x=pm1$, we obtain $f(-y)/f(-1)=f(y)/f(1)$ for all $y$. So either
        $f(-x)=f(x)$ for all $x$ or $f(-x)=-f(x)$ for all $x$.



        The solution $y$ of $x^2+y=xy$ is $y=frac{x^2}{x-1}$ unless $x=1$. Using this $y$ we have
        $$fleft(frac{x^3}{x-1}right)=frac{f(x)^3}{f(x)-1}mbox{ for }xneq1.$$
        Consider now a solution $z$ of $frac{z^2}{z-1}=-1$. Then $f(-z)=frac{f(z)^3}{f(z)-1}$.
        As a consequence $f(-z)$ cannot equal $f(z)$ because $1=frac{f(z)^2}{f(z)-1}$ has no real
        solution for $f(z)$. Hence $f(-z)=-f(z)$ and, as seen above, $f(-x)=-f(x)$ for all $x.$



        Next consider $x=1$ and replace $y$ by $-y$. We find $f(1-y)=f(1)^2-frac{f(y)}{f(1)}$
        for $yneq1$. Replacing here $y$ by $1-y$ yields with $w=f(1)$
        $$w^2f(y)=w^4-wf(1-y)=w^4-w^3+f(y)mbox{ for all }yin A,yneq1.$$
        Thus $(w^2-1)f(y)=w^4-w^3$. So we must have $w=1$ as otherwise $w=-1$
        which leads to a contradiction or $f(y)=frac{w^3}{w+1}$ is constant which had been excluded.
        So finally we proved that $w=f(1)=1$.



        Putting $x=1$ now gives $f(y+1)=f(y)+1$ for $yneq-1$. In particular, we find $f(n)=n$ and
        $f(-n)=-n$ for all positive integers $n$. Putting $x=n$, we obtain
        $f(y+n^2)=n^2+frac1nf(ny)$ for all $yneq-n^2$ and $f(ny)=nf(y)$ for all positive $y$
        (In fact, we only have to exclude that $y$ is a negative integer).
        In particular this leads to $f(q)=q$ for all rational positive $q$. It is extended to
        negative rationals using $f(-x)=-f(x)$.
        Moreover, we obtain $f(y+m)=f(y)+m$ for all non-integer $y$ and thus
        $f(y+frac mn)=frac1nf(ny+m)=frac1n(nf(y)+m)=f(y)+frac mn$
        for all irrational $y$ and hence $f(y+q)=f(y)+q$
        for all rational $q$ and irrational $y$.



        Finally, as $f(x^2+1)=f(x)^2+1$ and hence $f(x^2)=f(x)^2$ for all $x$,
        the values of $f$ for positive $y$ are positive and hence also its values for
        negative $y$ are negative.
        Given any irrational $y$, we consider rational $q,r$ with $q<y<r$. Then
        as $y-q>0$, we have $f(y)-q=f(y-q)>0$ and hence $q<f(y)$. In the same way we show that
        $f(y)<r$. Since we can use any rational $q,r$ with $q<y<r$,
        we have shown that $f(y)=y$ also for irrational $y$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 5 '18 at 9:28

























        answered Dec 5 '18 at 9:23









        HelmutHelmut

        664118




        664118






























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