Number Theory: Prove that $gcd(a,b) le sqrt{a+b}$












2












$begingroup$



For positive integers $a$ and $b$, we know $dfrac{a+1}{b} + dfrac{b+1}{a}$ is also
a positive integer. Prove that $gcd(a,b) le sqrt{a+b}$.




Using Bézout's lemma, we know that $gcd(a, b) = sa + tb$. I want to prove that $(sa+tb)^2 le a+b$. We know $ab,|,a(a+1) + b(b+1)$.



Therefore, $(sa + tb)^2 le (sa)^2 + (tb)^2 + 2st(a(a+1)+b(b+1))$.



I'm not sure how can continue from here. Any ideas to continue, or for a better way to prove the statement?



Thanks in advance.






elementary-number-theory discrete-mathematics







share|cite|improve this question











$endgroup$












  • $begingroup$
    Is it suppose to be $(sa+tb)^2 leq a+b$?
    $endgroup$
    – Larry
    Dec 7 '18 at 23:23










  • $begingroup$
    @Larry yes. fixed it.
    $endgroup$
    – Anthony
    Dec 7 '18 at 23:25










  • $begingroup$
    A related question (not a duplicate nor an answer): math.stackexchange.com/questions/1417404/…
    $endgroup$
    – Jean-Claude Arbaut
    Dec 7 '18 at 23:40


















2












$begingroup$



For positive integers $a$ and $b$, we know $dfrac{a+1}{b} + dfrac{b+1}{a}$ is also
a positive integer. Prove that $gcd(a,b) le sqrt{a+b}$.




Using Bézout's lemma, we know that $gcd(a, b) = sa + tb$. I want to prove that $(sa+tb)^2 le a+b$. We know $ab,|,a(a+1) + b(b+1)$.



Therefore, $(sa + tb)^2 le (sa)^2 + (tb)^2 + 2st(a(a+1)+b(b+1))$.



I'm not sure how can continue from here. Any ideas to continue, or for a better way to prove the statement?



Thanks in advance.






elementary-number-theory discrete-mathematics







share|cite|improve this question











$endgroup$












  • $begingroup$
    Is it suppose to be $(sa+tb)^2 leq a+b$?
    $endgroup$
    – Larry
    Dec 7 '18 at 23:23










  • $begingroup$
    @Larry yes. fixed it.
    $endgroup$
    – Anthony
    Dec 7 '18 at 23:25










  • $begingroup$
    A related question (not a duplicate nor an answer): math.stackexchange.com/questions/1417404/…
    $endgroup$
    – Jean-Claude Arbaut
    Dec 7 '18 at 23:40
















2












2








2


4



$begingroup$



For positive integers $a$ and $b$, we know $dfrac{a+1}{b} + dfrac{b+1}{a}$ is also
a positive integer. Prove that $gcd(a,b) le sqrt{a+b}$.




Using Bézout's lemma, we know that $gcd(a, b) = sa + tb$. I want to prove that $(sa+tb)^2 le a+b$. We know $ab,|,a(a+1) + b(b+1)$.



Therefore, $(sa + tb)^2 le (sa)^2 + (tb)^2 + 2st(a(a+1)+b(b+1))$.



I'm not sure how can continue from here. Any ideas to continue, or for a better way to prove the statement?



Thanks in advance.






elementary-number-theory discrete-mathematics







share|cite|improve this question











$endgroup$





For positive integers $a$ and $b$, we know $dfrac{a+1}{b} + dfrac{b+1}{a}$ is also
a positive integer. Prove that $gcd(a,b) le sqrt{a+b}$.




Using Bézout's lemma, we know that $gcd(a, b) = sa + tb$. I want to prove that $(sa+tb)^2 le a+b$. We know $ab,|,a(a+1) + b(b+1)$.



Therefore, $(sa + tb)^2 le (sa)^2 + (tb)^2 + 2st(a(a+1)+b(b+1))$.



I'm not sure how can continue from here. Any ideas to continue, or for a better way to prove the statement?



Thanks in advance.






elementary-number-theory discrete-mathematics




elementary-number-theory discrete-mathematics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 9 '18 at 17:00









amWhy

1




1










asked Dec 7 '18 at 23:08









AnthonyAnthony

324




324












  • $begingroup$
    Is it suppose to be $(sa+tb)^2 leq a+b$?
    $endgroup$
    – Larry
    Dec 7 '18 at 23:23










  • $begingroup$
    @Larry yes. fixed it.
    $endgroup$
    – Anthony
    Dec 7 '18 at 23:25










  • $begingroup$
    A related question (not a duplicate nor an answer): math.stackexchange.com/questions/1417404/…
    $endgroup$
    – Jean-Claude Arbaut
    Dec 7 '18 at 23:40




















  • $begingroup$
    Is it suppose to be $(sa+tb)^2 leq a+b$?
    $endgroup$
    – Larry
    Dec 7 '18 at 23:23










  • $begingroup$
    @Larry yes. fixed it.
    $endgroup$
    – Anthony
    Dec 7 '18 at 23:25










  • $begingroup$
    A related question (not a duplicate nor an answer): math.stackexchange.com/questions/1417404/…
    $endgroup$
    – Jean-Claude Arbaut
    Dec 7 '18 at 23:40


















$begingroup$
Is it suppose to be $(sa+tb)^2 leq a+b$?
$endgroup$
– Larry
Dec 7 '18 at 23:23




$begingroup$
Is it suppose to be $(sa+tb)^2 leq a+b$?
$endgroup$
– Larry
Dec 7 '18 at 23:23












$begingroup$
@Larry yes. fixed it.
$endgroup$
– Anthony
Dec 7 '18 at 23:25




$begingroup$
@Larry yes. fixed it.
$endgroup$
– Anthony
Dec 7 '18 at 23:25












$begingroup$
A related question (not a duplicate nor an answer): math.stackexchange.com/questions/1417404/…
$endgroup$
– Jean-Claude Arbaut
Dec 7 '18 at 23:40






$begingroup$
A related question (not a duplicate nor an answer): math.stackexchange.com/questions/1417404/…
$endgroup$
– Jean-Claude Arbaut
Dec 7 '18 at 23:40












2 Answers
2






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3












$begingroup$

Suppose $dmid a$, $dmid b$.
Then
$$n:=left(frac{a+1}b+frac{b+1}aright)cdot frac adcdot frac bd-frac{a^2}{d^2}-frac{b^2}{d^2}$$
is an integer. Conclude.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Care to elaborate?
    $endgroup$
    – RandomMathGuy
    Dec 7 '18 at 23:28



















4












$begingroup$

Hagen von Eitzen does not seem to want to explain his answer further, so I will attempt to do so here:



Suppose that $d|a$ and $d|b$. Then we have the following:
begin{align*}
frac{a+b}{d^2} &= frac{a^2+a}{d^2}+frac{b^2+b}{d^2}-frac{a^2}{d^2}-frac{b^2}{d^2} \ &=left(frac{a+1}{b}+frac{b+1}{a}right)cdotfrac{a}{d}cdotfrac{b}{d}-frac{a^2}{d^2}-frac{b^2}{d^2}
end{align*}



The RHS is an integer by the hypothesis of the problem. The LHS is positive by the hypothesis of the problem. Thus we can safely conclude that $frac{a+b}{d^2}geq 1$ for all common divisors of $a$ and $b$, which is what we wanted to show.






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

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    active

    oldest

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    active

    oldest

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    3












    $begingroup$

    Suppose $dmid a$, $dmid b$.
    Then
    $$n:=left(frac{a+1}b+frac{b+1}aright)cdot frac adcdot frac bd-frac{a^2}{d^2}-frac{b^2}{d^2}$$
    is an integer. Conclude.






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      Care to elaborate?
      $endgroup$
      – RandomMathGuy
      Dec 7 '18 at 23:28
















    3












    $begingroup$

    Suppose $dmid a$, $dmid b$.
    Then
    $$n:=left(frac{a+1}b+frac{b+1}aright)cdot frac adcdot frac bd-frac{a^2}{d^2}-frac{b^2}{d^2}$$
    is an integer. Conclude.






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      Care to elaborate?
      $endgroup$
      – RandomMathGuy
      Dec 7 '18 at 23:28














    3












    3








    3





    $begingroup$

    Suppose $dmid a$, $dmid b$.
    Then
    $$n:=left(frac{a+1}b+frac{b+1}aright)cdot frac adcdot frac bd-frac{a^2}{d^2}-frac{b^2}{d^2}$$
    is an integer. Conclude.






    share|cite|improve this answer









    $endgroup$



    Suppose $dmid a$, $dmid b$.
    Then
    $$n:=left(frac{a+1}b+frac{b+1}aright)cdot frac adcdot frac bd-frac{a^2}{d^2}-frac{b^2}{d^2}$$
    is an integer. Conclude.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 7 '18 at 23:25









    Hagen von EitzenHagen von Eitzen

    277k22269496




    277k22269496








    • 1




      $begingroup$
      Care to elaborate?
      $endgroup$
      – RandomMathGuy
      Dec 7 '18 at 23:28














    • 1




      $begingroup$
      Care to elaborate?
      $endgroup$
      – RandomMathGuy
      Dec 7 '18 at 23:28








    1




    1




    $begingroup$
    Care to elaborate?
    $endgroup$
    – RandomMathGuy
    Dec 7 '18 at 23:28




    $begingroup$
    Care to elaborate?
    $endgroup$
    – RandomMathGuy
    Dec 7 '18 at 23:28











    4












    $begingroup$

    Hagen von Eitzen does not seem to want to explain his answer further, so I will attempt to do so here:



    Suppose that $d|a$ and $d|b$. Then we have the following:
    begin{align*}
    frac{a+b}{d^2} &= frac{a^2+a}{d^2}+frac{b^2+b}{d^2}-frac{a^2}{d^2}-frac{b^2}{d^2} \ &=left(frac{a+1}{b}+frac{b+1}{a}right)cdotfrac{a}{d}cdotfrac{b}{d}-frac{a^2}{d^2}-frac{b^2}{d^2}
    end{align*}



    The RHS is an integer by the hypothesis of the problem. The LHS is positive by the hypothesis of the problem. Thus we can safely conclude that $frac{a+b}{d^2}geq 1$ for all common divisors of $a$ and $b$, which is what we wanted to show.






    share|cite|improve this answer









    $endgroup$


















      4












      $begingroup$

      Hagen von Eitzen does not seem to want to explain his answer further, so I will attempt to do so here:



      Suppose that $d|a$ and $d|b$. Then we have the following:
      begin{align*}
      frac{a+b}{d^2} &= frac{a^2+a}{d^2}+frac{b^2+b}{d^2}-frac{a^2}{d^2}-frac{b^2}{d^2} \ &=left(frac{a+1}{b}+frac{b+1}{a}right)cdotfrac{a}{d}cdotfrac{b}{d}-frac{a^2}{d^2}-frac{b^2}{d^2}
      end{align*}



      The RHS is an integer by the hypothesis of the problem. The LHS is positive by the hypothesis of the problem. Thus we can safely conclude that $frac{a+b}{d^2}geq 1$ for all common divisors of $a$ and $b$, which is what we wanted to show.






      share|cite|improve this answer









      $endgroup$
















        4












        4








        4





        $begingroup$

        Hagen von Eitzen does not seem to want to explain his answer further, so I will attempt to do so here:



        Suppose that $d|a$ and $d|b$. Then we have the following:
        begin{align*}
        frac{a+b}{d^2} &= frac{a^2+a}{d^2}+frac{b^2+b}{d^2}-frac{a^2}{d^2}-frac{b^2}{d^2} \ &=left(frac{a+1}{b}+frac{b+1}{a}right)cdotfrac{a}{d}cdotfrac{b}{d}-frac{a^2}{d^2}-frac{b^2}{d^2}
        end{align*}



        The RHS is an integer by the hypothesis of the problem. The LHS is positive by the hypothesis of the problem. Thus we can safely conclude that $frac{a+b}{d^2}geq 1$ for all common divisors of $a$ and $b$, which is what we wanted to show.






        share|cite|improve this answer









        $endgroup$



        Hagen von Eitzen does not seem to want to explain his answer further, so I will attempt to do so here:



        Suppose that $d|a$ and $d|b$. Then we have the following:
        begin{align*}
        frac{a+b}{d^2} &= frac{a^2+a}{d^2}+frac{b^2+b}{d^2}-frac{a^2}{d^2}-frac{b^2}{d^2} \ &=left(frac{a+1}{b}+frac{b+1}{a}right)cdotfrac{a}{d}cdotfrac{b}{d}-frac{a^2}{d^2}-frac{b^2}{d^2}
        end{align*}



        The RHS is an integer by the hypothesis of the problem. The LHS is positive by the hypothesis of the problem. Thus we can safely conclude that $frac{a+b}{d^2}geq 1$ for all common divisors of $a$ and $b$, which is what we wanted to show.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 7 '18 at 23:54









        RandomMathGuyRandomMathGuy

        1872




        1872






























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