8
$\begingroup$

Show that $x^2 + y^2$ and $x^2 - y^2$ cannot both be perfect squares at the same time where $x, y \in \mathbb{Z}^+$.

I think that $x^2 + 2xy + y^2$ and $x^2 + y^2$ are not consecutive squares since the difference is even. I think it has some relation with other squares like $(x+y)^2$ and $(x-y)^2$.

How should I proceed? I would love some hints.

$\endgroup$
  • $\begingroup$ I'm sure there was a question in MSE that $a+b $ and $a+2b$ cannot be squares at the same time. $\endgroup$ – N.S.JOHN Apr 22 '16 at 14:41
  • $\begingroup$ what is perfect square? $\endgroup$ – SiXUlm Apr 22 '16 at 14:42
  • $\begingroup$ @sixulm Squares of integers. $\endgroup$ – N.S.JOHN Apr 22 '16 at 14:43
  • $\begingroup$ @N.S.JOHN I see, so it's simply square. $\endgroup$ – SiXUlm Apr 22 '16 at 14:43
  • $\begingroup$ I believe it means x and y are both non-zero integers. $\endgroup$ – Mc Cheng Apr 22 '16 at 14:51
12
$\begingroup$

Suppose $x^2+y^2=a^2$ and $x^2-y^2=b^2$, then by multiplying them you get $x^4-y^4=(ab)^2$. This last equation has no non-trivial solution; see e.g. Solving $x^4-y^4=z^2$.

| cite | improve this answer | |
$\endgroup$
  • 2
    $\begingroup$ This proves a stronger result - that $x^2+y^2$ and $x^2-y^2$ can't have their product equal to a square (except for trivial cases). $\endgroup$ – user236182 Apr 22 '16 at 14:57
  • $\begingroup$ @user236182 Exactly. $\endgroup$ – user148212 Apr 22 '16 at 15:03
  • $\begingroup$ I think that this proof can be modified for Fermat's Last Theorem where $n=4$. This question itself can be a different form of asking for a proof of the FLT where $n=4$. No doubt my problem sheet says this question was asked by Fermat. $\endgroup$ – TheRandomGuy Apr 23 '16 at 13:50
0
$\begingroup$

Hint:

If$ x^2+y^2=c^2$ is a Pythagorean triple than there are two integers $m,n$ such that: $$ x=m^2-n^2 \qquad y=2mn \qquad c=m^2+n^2 $$

so: $$ x^2-y^2=(m^2-n^2)^2-4m^2n^2=m^4+n^4-6 m^2n^2 $$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Provided $x$ and $y$ are coprime, but this is not an essential restriction. You should also justify why $x$ is odd and $y$ is even (after reduction to a primitive Pythagorean triple). $\endgroup$ – egreg Apr 22 '16 at 15:36
  • 2
    $\begingroup$ Why can't $m^4+n^4-6 m^2n^2$ be a perfect square? $\endgroup$ – TheRandomGuy Apr 22 '16 at 15:38
  • $\begingroup$ Let m^2=x & n^2=y Then the equation reduces to x^2-6xy+y^2 In which the discriminant is 0 if and only if y=8 or n=2\sqrt 2 So this would not have any integer solution $\endgroup$ – Atul Mishra Dec 8 '16 at 6:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.