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