Htykuut

i was looking for a function that satisfies $f(a+b) =frac{f(a)+f(b)}{f(a)*f(b)}$ for all $a,b in Bbb{N}$. I have never seen such a problem before and i would like some kind of help to get me started.

Asume that $f$ is defined on ${mathbb N}_{geq1}$, and let $f(1):=c$ for a $cindot{mathbb C}$ to be determined. From
$$f(n+1)={1over c}+{1over f(n)}qquad(ngeq1)$$
we compute
$$bigl(f(n)bigr)_{1leq nleq 5}=left(c,{2over c},{2+c^2over 2c},{2+3c^2over c(2+c^2)},{2+5c^2+c^4over c(2+3c^2)}right) .tag{1}$$
Now we should have
$$f(4)=f(2+2)={2 f(2)over bigl(f(2)bigr)^2}={2over f(2)}=c .$$
Comparing this with the value of $f(4)$ in $(1)$ we obtain the equation
$$c={2+3c^2over c(2+c^2)}$$
with the solutions $pm i$ and $pmsqrt{2}$. Starting anew with
$$f(5)=f(2+3)=ldots ,$$
and comparing with the $f(5)$ from the table $(1)$ leads again to $c=pmsqrt{2}$ and to four complex $c$-values $nepm i$. It follows that necessarily $cin{pmsqrt{2}}$. It is then easily checked that the function $f(n):=c$ for all $ngeq1$ satisfies the given functional equation.

It is not always clear whether $mathbb N$ includes $0$ or not. If we ignore zero, then one way of working is to set $f(1)=x$ whence:
$$f(2)=f(1+1)=frac 2x$$
$$f(3)=f(2+1)=frac {x^2+2}{2x}$$
$$f(4)=f(3+1)=frac {3x^2+2}{x^3+2x}$$but also $$f(4)=f(2+2)=x$$

In fact we have $f(2a)=f(a+a)=frac 2{f(a)}$ so that $f(a)f(2a)=2=f(2a)f(4a)$ so that always $f(4a)=f(a)$

Equating the expressions for $f(4)$ gives an equation which can be solved for possible values of $x$

Assuming $0in mathbb{N}$ we get $f(0)=pm sqrt 2$ as noted in comments.

1. Let $f(0)=sqrt 2$
   

Then $f(1)=frac{f(1)+f(0)}{f(1)cdot f(0)}=frac{f(1)+sqrt 2}{sqrt 2 cdot f(1)}$ leads to the quadratic equation $sqrt 2 cdot [f(1)]^2-f(1)-sqrt 2 =0$ wich has solutions $f(1)=sqrt 2;$ and $;f(1)=-frac{sqrt 2}{2}.$

• $f(1)=sqrt 2;$ leads necessarily to $$;f(a)=sqrt 2quadtext{for}quad ain mathbb{N},$$ because the sequence defined as $a_1=sqrt 2,; a_{n+1}=frac{a_n+sqrt 2}{sqrt 2cdot a_n},; nin mathbb{N};$ is constant.




• Consider $f(1)=-frac{sqrt 2}{2}.$ We get $$f(2)=frac{2}{f(1)}=-2sqrt2,quad f(3)=frac{f(2)+f(1)}{f(2)f(1)}=-frac{5sqrt 2}{4},$$ but then we would obtain two different values of $f(4):$ $$f(4)=frac{2}{f(2)}=-frac{sqrt2}{2}quad text{or}quad f(4)=frac{f(3)+f(1)}{f(3)f(1)}=-frac{7sqrt2}{5}$$

Therefore $f(1)=sqrt2$ and the function is constant.

2. 
   

   Let $f(0)=-sqrt 2$

   
   
   

   Similarly to the above, $f(1)$ can take two values.

   
   
   
   
   
   • 
     $f(1)=-sqrt2$ determines a constant solution $f(n)=-sqrt2,$
   
   
   • the other is $f(1)=frac{sqrt2}{2}$ and leads to a disambiguation.
   
   
   
   

Conclusion

The only solutions defined in $0$ are the constant functions $f(n)=sqrt2;$ and $g(n)=-sqrt 2, n=0,1,;dots$