MHB Prove: If an Integer is a Sum of Squares, its Square is Also a Sum of Squares

[anemone]

Here is this week's POTW:

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Suppose $a$ is an integer that is a sum of squares of three positive integers. Prove that $a^2$ is also a sum of squares of three positive integers.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

No one answered last week problem. :(

You can see the proposed solution as follows:

Spoiler

Let $a=p^2+q^2+r^2$.

Therefore

$\begin{align*}a^2&=(p^2+q^2+r^2)^2\\&=p^4+q^4+r^4+2p^2q^2+2q^2r^2+2r^2p^2\\&=(p^4+q^4+r^4+2p^2q^2-2q^2r^2-2r^2p^2)+4q^2r^2+4r^2p^2\\&=(p^2+q^2-r^2)^2+(2pr)^2+(2qr)^2\end{align*}$

Without loss of generality, we may assume $p^2 \ge q^2\ge r^2$, so we have $p^2+q^2-r^2\ge 0$ and this completes the proof.