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. – Bill Dubuque Feb 04 '17 at 20:08
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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 – user25406 Feb 04 '17 at 20:22
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http://math.stackexchange.com/questions/1127654/parametrization-of-solutions-of-diophantine-equation – individ Feb 05 '17 at 11:45
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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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2@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. – Erick Wong Feb 04 '17 at 20:21