Is there a proof that doubling the sum of squares of any two integers will always equate to another sum of two squares?

Question

How do I prove:
$$\forall \space a,b \in \mathbb Z, \exists \space c,d \in \mathbb Z: 2*(a^2+b^2) = c^2 + d^2$$ $$d<>a, d<> b$$

For example, : $$2*(1^2 + 2^2) = 1^2 + 3^2 = 10$$ $$2*(2^2 + 3^2) = 1^2 + 5^2 = 26$$ $$2*(2^2 + 4^2) = 2^2 + 6^2 = 40$$ $$etc$$

Multiplying by a square number would be trivial, but are there any other non-square multiples other than two where this may also be true?

There’s a standard result that an integer $>1$ is a sum of two squares iff its prime factorization contains no prime $p$ congruent to $3$ mod $4$ with odd multiplicity. — MathematicsStudent1122, Jan 10 '20 at 02:13

score 14 · Accepted Answer · answered Jan 10 '20 at 02:22

14

$2(a^2+b^2)=(a+b)^2+(a-b)^2$.

answered Jan 10 '20 at 02:22

Oscar Lanzi

1

Many thanks - this is beautifully simple. – Neil Ennis Jan 10 '20 at 02:25
1

That's what we're here for. – Oscar Lanzi Jan 10 '20 at 02:27
Maybe noteworthy: This even proves the result for $\mathbb Z^+$ as long as $a\neq b$. – Torsten Schoeneberg Jan 10 '20 at 03:03
3

More generally: $(a^2 + b^2)(c^2 + d^2) = (ad+bc)^2 + (ac-bd)^2$. – Robert Israel Jan 10 '20 at 03:31

1 Answers1