I am currently learning direct proofs. I couldn't solve the following exercise.

Define an integer $m$ to be bisquare iff, $\exists a \in Z, \exists b \in Z, m = a^2 + b^2$.

Let $r$ and $s$ be fixed integers.

Prove: If $r$ and $s$ are bisquare, then $rs$ is bisquare.

My work:

Proof. Assume $r$ and $s$ are bisquare. I must prove that $rs$ is bisquare. I have assumed that $\exists a \in Z, \exists b \in Z, r = a^2 + b^2$ and $\exists x \in Z, \exists y \in Z, s = x^2 + y^2$. I must prove that $\exists m \in Z, \exists n \in Z, rs = m^2 + n^2$.

From my assumption, I can show that

$rs = (a^2+b^2)(x^2+y^2)$.

This is where I got stuck. I've tried a couple algebraic tricks but I didn't get anywhere.

  • 1
    $\begingroup$ Let's at least note that you've gotten all of the logical structure correct and know that you're looking for some creative algebraic step to carry on—that's a good start! $\endgroup$ Jun 26 at 17:03

1 Answer 1


Note that: $$(a^2+b^2)(x^2+y^2)=a^2x^2+a^2y^2+b^2x^2+b^2y^2$$ We would like to write this as sum of squares. Let's try: $$a^2x^2+b^2y^2=a^2x^2+b^2y^2+2abxy-2abxy=(ax+by)^2-2abxy$$ $$a^2y^2+b^2x^2=a^2y^2+b^2x^2+2abxy-2abxy=(ay-bx)^2+2abxy$$ Therefore: $$(a^2+b^2)(x^2+y^2)=a^2x^2+b^2y^2+a^2y^2+b^2x^2+2abxy-2abxy=(ax+by)^2+(ay-bx)^2$$

  • $\begingroup$ Alongside this calculation it's always good for us to record (for present or future enlightenment) the equivalent identity $|a+bi||y+ix| = |(a+bi)(y+ix)|$ in the complex numbers. $\endgroup$ Jun 26 at 17:05

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