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For all $a,b\in\mathbb Z$ there are $c,d\in\mathbb Z$ such that $2(a^2+b^2)=c^2+d^2$.

The conjecture is tested for $a^2+b^2<1,000,000$ but I have problems with proving it.

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All you need is this: $$2(a^2+b^2)=(a+b)^2+(a-b)^2$$

which is a special case of Diophantus' identity

$$(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2$$

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    $\begingroup$ @Lehs A strong consequence of the general identity is that the set of sums of two squares is closed under multiplication: not only does it work for two times the sum of two squares, it also works for 5 times, 10 times, 13 times, etc. $\endgroup$ – Erick Wong Feb 4 '17 at 20:21

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