I was trying to do this proof where:

Assume $a,b,r,s$ are relatively prime, and that $$a^2+b^2=r^2$$ and $$a^2-b^2=s^2$$ Prove that $a,r,s$ are odd and $b$ is even.

So I started off by saying that if a number $x^2$ is odd, then $x$ is odd. Same applies for even numbers. Then I said that both $a$ and $b$ can't be even at the same time, and can't be odd, since if they are both odd, $r$ and $s$ will both be even. But then $b$ can be odd while $a$ is even and vice versa: both $r$ and $s$ will be odd. Am I missing something?


  • 1
    $\begingroup$ Prove $b$ is even first, by looking at the equation modulo $4$ and seeing that $r^2-2b^2=s^2$. $\endgroup$ – Thomas Andrews Mar 31 '15 at 1:04

So as you say, $a,b$ can't both be even (given that they're relatively prime) and they can't both be odd, since then $r,s$ would both be even (but must be relatively prime).

Then exactly one of $a,b$ is even. Up until here you did well, but next $a$ can't be even with $b$ odd, since then $s$ is odd, but $$s^2+b^2=(2s_1+1)^2+(2b_1+1)^2=4(s_1^2+s_1+b_1^2+b_1)+2\neq a^2=(2a_1)^2=4a_1^2$$ because of divisibility by $4$.

Thus $b$ is even, $a$ is odd. So $s$ is odd and $r$ is odd.


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.