I ask a question that is probably never proved a conjecture: if $x$ and $y$ are two natural numbers $> 1$ such that $2xy = N^2$ ( double their product is a perfect square ), x and y cannot be part of a Pythagorean triple , that is, there is no natural $M$, $M > 1$ such that $x^2 + y^2 = M^2$.
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If so then $,(x-y)^2,,x^2+y^2,,(x+y)^2,$ are in arithmetic progression, with square step size $,2xy,,$ so Fibonacci's Lost Theorem applies, cf. linked dupe. – Bill Dubuque Aug 31 '24 at 06:46
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This conjecture is true.
Assume for contradiction such $x$, $y$ exist. We may assume that $\gcd(x,y) = 1$ without loss of generality. Then according to $x^2+y^2=M^2$ there exist $m$ and $n$ of different parity and $\gcd(m,n) = 1$ such that $x = m^2-n^2$ and $y = 2mn$ (Euclid's Formula), so we would need $mn(m-n)(m+n)$ to be a square.
As $m$, $n$, $m-n$, $m+n$ are all pairwise coprime (note exactly one of them is odd), this can only happen if each is a perfect square; if $m = a^2$, $n = b^2$ then we get $c^2 = a^4-b^4$, but this is not possible.
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2@EnzoAlberio Note that $2yx$ is a square, that is $2(2mn)(m^2-n^2)=4mn(m^2-n^2)$ is a square. Therefore we can divide by 4 to see that $mn(m^2-n^2)$ is a square. – Oct 28 '15 at 12:03