Problem: If $q$ is an integer that can be expressed as the sum of integer squares,show that both $2q$ and $5q$ can also be espressed as the sum of two integer squares.
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4This has nothing to do with the pythagorean theorem. What if $q = 1 + 4$? What is $c$? – Umberto P. Sep 22 '15 at 17:21
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1$q$ is not given to be a square, so $c$ may not be an integer. – Robert Israel Sep 22 '15 at 17:25
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thanks for your comments, i have missed it all up, let me edit. – Nameless Sep 22 '15 at 17:27
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one question : in general why can't i suppose $q$ to be the square of some number which squared does give me an integer (even if that number isn't an integer)? – Nameless Sep 22 '15 at 18:13
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If $q=a^2+b^2$, then $(a+b)^2+(a-b)^2=2q$.
For $5q$, use (a special case of) Brahmagupta's identity: $$(x^2+y^2)(a^2+b^2)=(ax-by)^2+(ay+bx)^2$$
Bernard
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For $5q$, we can write it as
$$(a+2b)^2 + (2a-b)^2=a^2+4ab+4b^2+4a^2-4ab+b^2$$ $$=5a^2+5b^2=5 (a^2+b^2)=5 q$$
We only need to prove that $a+2b$ and $2a-b$ are integers.
QC_QAOA
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Larry Zheng
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