2
$\begingroup$

Supposed $a,b \in \mathbb{Z}$. If $ab$ is odd, then $a^{2} + b^{2}$ is even.

I'm stuck on the best way to get this started. My thinking is that I could use cases. i.e.

  • Case 1: a is even and b is odd
  • Case 2: a is odd and b is even
  • Case 3: a is odd and b is odd

Would this be my best approach? Or is there an easier way to look at it? Thanks.

$\endgroup$
0
5
$\begingroup$

If $ab$ is odd then $a$ and $b$ must be both odd and then so are $a^2$ and $b^2$. But $a^2+b^2$ is the sum of two odd numbers, so it must be even.

$\endgroup$
1
  • 1
    $\begingroup$ Ah, this makes sense. It was a lot easier than what I was making it out to be. Thanks a lot! $\endgroup$ – Harry Oct 26 '14 at 20:18
1
$\begingroup$

$ab$ is odd iff a and b are both odd. That means $a^2$ and $b^2$ are both odd. The sum of two odd numbers is even. Therefore, $a^2+b^2$ must be even if $ab$ is odd.

$\endgroup$
0
$\begingroup$

More generally note: $\ ab(a^2+b^2)\,$ is even since

${\rm mod}\ 2\!:\ x^2\equiv x\,$ $\,\Rightarrow\,ab(a^2+b^2)\equiv ab(a+b)\equiv ab+ab\equiv 2ab\equiv 0$

$\endgroup$

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.