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

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.

• 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! Jun 26 at 17:03

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: $$(a^2+b^2)(x^2+y^2)=a^2x^2+b^2y^2+a^2y^2+b^2x^2+2abxy-2abxy=(ax+by)^2+(ay-bx)^2$$
• 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. Jun 26 at 17:05