If $\gcd(a,b,c)=1$ and $a^2 + b^2 = c^2$, prove $12|abc$ I know from $a^2+b^2=c^2$ that $a = st$, b = $\frac{s^2-t^2}{2}$, c = $\frac{s^2+t^2}{2}$ , where $s>t>1$, where $s$ and $t$ are both odd, but how do I make use of this information? 

$s,t$ odd means $s^2\equiv t^2\equiv 1\mod 8$, hence $b$ is divisible by $4$. If neither $s$ nor $t$ is divisible by $3$, then $s^2\equiv t^2\equiv 1\mod 3$ and $3|b$. Otherwise $3|a$, in both cases $12|abc$.
