Prove that $a$, $r$, and $s$ are odd and $b$ is even Question: Suppose that $a$, $b$, $r$, $s$ are pairwise relatively prime and that $a^2 + b^2 = r^2$ and $a^2 - b^2 = s^2$. Prove that $a$, $r$, and $s$ are odd and $b$ is even.
My incomplete answer: Suppose $r$ is even then $a$ and $b$ are both even or both odd; then $s$ even, which is a contradiction with the assumption that $\gcd(r, s) = 1$. Therefore, $r$ is odd, then one of $a$ and $b$ is odd and the other is even, thus $s$ is odd. Since $\gcd(r, s) = 1$, then $|r - s| \ge 2$ (which doesn't help to solve!). How to know which of $a$ or $b$ is odd (then the other will be even)?
Thank you.      

Use the fact that if $x$ is odd, then $x^2 \equiv 1 \pmod{4}$. This is because $(2t + 1)^2 = 4(t^2 + t) + 1$.
Now first look at $a^2 + b^2 = r^2$. If $r$ is even, then $r^2 \equiv 0 \pmod{4}$. Then either (i) $a$ and $b$ are both even or (ii) $a$ and $b$ are both odd. In Case (i), we violate the gcd condition. In Case (ii), $a^2 + b^2 \equiv 2 \pmod{4}$, contradicting the fact that $r^2 \equiv 0 \pmod{4}$.
So $r$ is odd, and one of $a$ or $b$ is odd, and the other is even. We use $a^2 - b^2 = s^2$ to show $a$ must be odd. For if $a$ is even, then $a^2 - b^2 \equiv -1 \pmod{4}$, contradicting the fact that $s^2 \equiv 1 \pmod{4}$.
