0
$\begingroup$

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.

$\endgroup$
1
  • 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

3
$\begingroup$

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: $$(a^2+b^2)(x^2+y^2)=a^2x^2+b^2y^2+a^2y^2+b^2x^2+2abxy-2abxy=(ax+by)^2+(ay-bx)^2$$

$\endgroup$
1
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.