When $(m+n)(m-n)$ is perfect square

Question

If $2xy$ is a perfect square, then $x^2+y^2$ cannot be
I was trying to prove this way -
Let assume $2xy = n^2$ and $x^2+y^2 = m^2$ exists. Now -
$x^2+y^2 - 2xy = (m+n)(m-n)\\ (x-y)^2 = (m+n)(m-n)$
How do I continue to prove that such $m$ or such $x,y$ doesn't exists?

$(x-y,n,m)$ would be a Pythagorean triple. – Jonas Meyer Feb 02 '17 at 02:42 — Jonas Meyer, Feb 02 '17 at 02:42
Where did you see this question? – N.S.JOHN Feb 02 '17 at 14:08 — N.S.JOHN, Feb 02 '17 at 14:08
@N.S.JOHN Yesterday :| – Rezwan Arefin Feb 02 '17 at 14:22 — Rezwan Arefin, Feb 02 '17 at 14:22
@N.S.JOHN I was just browsing MSE .. and then find it :) – Rezwan Arefin Feb 02 '17 at 14:30 — Rezwan Arefin, Feb 02 '17 at 14:30

score 1 · Accepted Answer · answered Feb 02 '17 at 03:19

If both $2xy$ and $x^2+y^2$ are squares, then since $(x^2-y^2)^2+(2xy)^2=(x^2+y^2)^2,$ there would be two fourth powers whose difference is a square, known impossible except for trivial cases. [Technically $x=y=0$ works, but I'm assuming you want solutions in nonzero integers.]

Note: that impossibility proof isn't easy...

When $(m+n)(m-n)$ is perfect square

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