Find integer solutions to the equation $2^x+3^y=z^2$












2












$begingroup$


Here's my approach:



Since $2^x$ is always even we can rewrite it as $2k$,
and similarly $3^y$ is always odd so we can write it as $2m+1$.
The sum is equated to a square so $z^2=4r+1$ (it is always an odd square since odd +even is always odd) but after this I do not know what to do.






number-theory







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    First thing that comes to my mind is difference of squares: If $x$ is even, say $x=2k$, then $3^y=z^2-2^{2k}=(z+2^k)(z-2^k)$ and it becomes a factoring problem. Using a modulo $3$ analysis, you can argue that $x$ must be even for $y>0$.
    $endgroup$
    – SmileyCraft
    Dec 12 '18 at 17:06








  • 1




    $begingroup$
    Please use mathjax the next time you write math.
    $endgroup$
    – harshit54
    Dec 12 '18 at 17:07
















2












$begingroup$


Here's my approach:



Since $2^x$ is always even we can rewrite it as $2k$,
and similarly $3^y$ is always odd so we can write it as $2m+1$.
The sum is equated to a square so $z^2=4r+1$ (it is always an odd square since odd +even is always odd) but after this I do not know what to do.






number-theory







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    First thing that comes to my mind is difference of squares: If $x$ is even, say $x=2k$, then $3^y=z^2-2^{2k}=(z+2^k)(z-2^k)$ and it becomes a factoring problem. Using a modulo $3$ analysis, you can argue that $x$ must be even for $y>0$.
    $endgroup$
    – SmileyCraft
    Dec 12 '18 at 17:06








  • 1




    $begingroup$
    Please use mathjax the next time you write math.
    $endgroup$
    – harshit54
    Dec 12 '18 at 17:07














2












2








2





$begingroup$


Here's my approach:



Since $2^x$ is always even we can rewrite it as $2k$,
and similarly $3^y$ is always odd so we can write it as $2m+1$.
The sum is equated to a square so $z^2=4r+1$ (it is always an odd square since odd +even is always odd) but after this I do not know what to do.






number-theory







share|cite|improve this question











$endgroup$




Here's my approach:



Since $2^x$ is always even we can rewrite it as $2k$,
and similarly $3^y$ is always odd so we can write it as $2m+1$.
The sum is equated to a square so $z^2=4r+1$ (it is always an odd square since odd +even is always odd) but after this I do not know what to do.






number-theory




number-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 12 '18 at 17:08









eyeballfrog

6,228629




6,228629










asked Dec 12 '18 at 16:59









mtommtom

254




254








  • 2




    $begingroup$
    First thing that comes to my mind is difference of squares: If $x$ is even, say $x=2k$, then $3^y=z^2-2^{2k}=(z+2^k)(z-2^k)$ and it becomes a factoring problem. Using a modulo $3$ analysis, you can argue that $x$ must be even for $y>0$.
    $endgroup$
    – SmileyCraft
    Dec 12 '18 at 17:06








  • 1




    $begingroup$
    Please use mathjax the next time you write math.
    $endgroup$
    – harshit54
    Dec 12 '18 at 17:07














  • 2




    $begingroup$
    First thing that comes to my mind is difference of squares: If $x$ is even, say $x=2k$, then $3^y=z^2-2^{2k}=(z+2^k)(z-2^k)$ and it becomes a factoring problem. Using a modulo $3$ analysis, you can argue that $x$ must be even for $y>0$.
    $endgroup$
    – SmileyCraft
    Dec 12 '18 at 17:06








  • 1




    $begingroup$
    Please use mathjax the next time you write math.
    $endgroup$
    – harshit54
    Dec 12 '18 at 17:07








2




2




$begingroup$
First thing that comes to my mind is difference of squares: If $x$ is even, say $x=2k$, then $3^y=z^2-2^{2k}=(z+2^k)(z-2^k)$ and it becomes a factoring problem. Using a modulo $3$ analysis, you can argue that $x$ must be even for $y>0$.
$endgroup$
– SmileyCraft
Dec 12 '18 at 17:06






$begingroup$
First thing that comes to my mind is difference of squares: If $x$ is even, say $x=2k$, then $3^y=z^2-2^{2k}=(z+2^k)(z-2^k)$ and it becomes a factoring problem. Using a modulo $3$ analysis, you can argue that $x$ must be even for $y>0$.
$endgroup$
– SmileyCraft
Dec 12 '18 at 17:06






1




1




$begingroup$
Please use mathjax the next time you write math.
$endgroup$
– harshit54
Dec 12 '18 at 17:07




$begingroup$
Please use mathjax the next time you write math.
$endgroup$
– harshit54
Dec 12 '18 at 17:07










2 Answers
2






active

oldest

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1












$begingroup$

Hint: Say $y>0$ then $3mid 3^y$, using $pmod 3$ we see that $x$ must be even so $x=2t$ and now we have $$3^y = (z-2^t)(z+2^t)$$



Now $z-2^t = 3^m$ and $z+2^t = 3^n$ so $$2^{t+1} = 3^m+3^n = 3^m(1+3^{n-m})$$



and so $m=0$ and so on...






share|cite|improve this answer











$endgroup$









  • 2




    $begingroup$
    Watch out for $2^3+3^0=3^2$ though.
    $endgroup$
    – SmileyCraft
    Dec 12 '18 at 17:09



















1












$begingroup$

First, deal with the small cases: $x=0, x=1$ or $y leq 2$.
Reduce the equation modulo some numbers to infer some properties of $x$ or $y$, eg if $x>1$, $y$ is even. Then you can factor $2^x=z^2-3^y$, and since it is a power of two the smallest factor must divide the other and must divide $2^x$. Then it is up to you.






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Hint: Say $y>0$ then $3mid 3^y$, using $pmod 3$ we see that $x$ must be even so $x=2t$ and now we have $$3^y = (z-2^t)(z+2^t)$$



    Now $z-2^t = 3^m$ and $z+2^t = 3^n$ so $$2^{t+1} = 3^m+3^n = 3^m(1+3^{n-m})$$



    and so $m=0$ and so on...






    share|cite|improve this answer











    $endgroup$









    • 2




      $begingroup$
      Watch out for $2^3+3^0=3^2$ though.
      $endgroup$
      – SmileyCraft
      Dec 12 '18 at 17:09
















    1












    $begingroup$

    Hint: Say $y>0$ then $3mid 3^y$, using $pmod 3$ we see that $x$ must be even so $x=2t$ and now we have $$3^y = (z-2^t)(z+2^t)$$



    Now $z-2^t = 3^m$ and $z+2^t = 3^n$ so $$2^{t+1} = 3^m+3^n = 3^m(1+3^{n-m})$$



    and so $m=0$ and so on...






    share|cite|improve this answer











    $endgroup$









    • 2




      $begingroup$
      Watch out for $2^3+3^0=3^2$ though.
      $endgroup$
      – SmileyCraft
      Dec 12 '18 at 17:09














    1












    1








    1





    $begingroup$

    Hint: Say $y>0$ then $3mid 3^y$, using $pmod 3$ we see that $x$ must be even so $x=2t$ and now we have $$3^y = (z-2^t)(z+2^t)$$



    Now $z-2^t = 3^m$ and $z+2^t = 3^n$ so $$2^{t+1} = 3^m+3^n = 3^m(1+3^{n-m})$$



    and so $m=0$ and so on...






    share|cite|improve this answer











    $endgroup$



    Hint: Say $y>0$ then $3mid 3^y$, using $pmod 3$ we see that $x$ must be even so $x=2t$ and now we have $$3^y = (z-2^t)(z+2^t)$$



    Now $z-2^t = 3^m$ and $z+2^t = 3^n$ so $$2^{t+1} = 3^m+3^n = 3^m(1+3^{n-m})$$



    and so $m=0$ and so on...







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 12 '18 at 17:10

























    answered Dec 12 '18 at 17:08









    greedoidgreedoid

    44.2k1155110




    44.2k1155110








    • 2




      $begingroup$
      Watch out for $2^3+3^0=3^2$ though.
      $endgroup$
      – SmileyCraft
      Dec 12 '18 at 17:09














    • 2




      $begingroup$
      Watch out for $2^3+3^0=3^2$ though.
      $endgroup$
      – SmileyCraft
      Dec 12 '18 at 17:09








    2




    2




    $begingroup$
    Watch out for $2^3+3^0=3^2$ though.
    $endgroup$
    – SmileyCraft
    Dec 12 '18 at 17:09




    $begingroup$
    Watch out for $2^3+3^0=3^2$ though.
    $endgroup$
    – SmileyCraft
    Dec 12 '18 at 17:09











    1












    $begingroup$

    First, deal with the small cases: $x=0, x=1$ or $y leq 2$.
    Reduce the equation modulo some numbers to infer some properties of $x$ or $y$, eg if $x>1$, $y$ is even. Then you can factor $2^x=z^2-3^y$, and since it is a power of two the smallest factor must divide the other and must divide $2^x$. Then it is up to you.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      First, deal with the small cases: $x=0, x=1$ or $y leq 2$.
      Reduce the equation modulo some numbers to infer some properties of $x$ or $y$, eg if $x>1$, $y$ is even. Then you can factor $2^x=z^2-3^y$, and since it is a power of two the smallest factor must divide the other and must divide $2^x$. Then it is up to you.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        First, deal with the small cases: $x=0, x=1$ or $y leq 2$.
        Reduce the equation modulo some numbers to infer some properties of $x$ or $y$, eg if $x>1$, $y$ is even. Then you can factor $2^x=z^2-3^y$, and since it is a power of two the smallest factor must divide the other and must divide $2^x$. Then it is up to you.






        share|cite|improve this answer









        $endgroup$



        First, deal with the small cases: $x=0, x=1$ or $y leq 2$.
        Reduce the equation modulo some numbers to infer some properties of $x$ or $y$, eg if $x>1$, $y$ is even. Then you can factor $2^x=z^2-3^y$, and since it is a power of two the smallest factor must divide the other and must divide $2^x$. Then it is up to you.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 12 '18 at 17:11









        MindlackMindlack

        4,740210




        4,740210






























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