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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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    See also [Show that every power of 2 is sum of two squares.](http://math.stackexchange.com/q/100506/242) – 2017-02-04
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    If you look at the algorithm below, you would see that one can calculate the sum of two squares of an integer N by calculating first the sum of two squares of the integer 2N. http://math.stackexchange.com/questions/2118459/sum-of-two-squares-of-an-integer-n-the-simplest-algorithm?noredirect=1#comment4356705_2118459 – 2017-02-04
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    http://math.stackexchange.com/questions/1127654/parametrization-of-solutions-of-diophantine-equation – 2017-02-05

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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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    @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. – 2017-02-04