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?
When $(m+n)(m-n)$ is perfect square
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elementary-number-theory
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0$(x-y,n,m)$ would be a Pythagorean triple. – 2017-02-02
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0Where did you see this question? – 2017-02-02
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0@N.S.JOHN Yesterday :| – 2017-02-02
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0Okay. I asked where. – 2017-02-02
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1@N.S.JOHN I was just browsing MSE .. and then find it :) – 2017-02-02
1 Answers
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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...