Two times the sum of two squares is a sum of two squares

Question

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.

Answer 1

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

which is a special case of Diophantus' identity

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