Show that for any prime numbers $p,q,r$, one has $p^2+q^2 \ne r^2$. Not sure how to start. Prove by induction? Please help!
 A: If both $p$ and $q$ are even, then $p=q=2$ and then $r^2=8$, not possible.
If both $p$ and $q$ are odd, then $r$ must be even and so $r=2$. But then $p,q\ge 3$ and so $p^2+q^2\ge 18 > 4=r^2$.
If $p$ is even and $q$ is odd, then $r$ is odd and $p=2$ and so $4=(r-q)(r+q)$. This implies $r-q=2$ and $r+q=2$, and so $q=0$, not possible; or $r-q=1$ and $r$ is even, not possible.
A: Since $p^2 = r^2 - q^2 = (r+q)(r-q)$ we reach an easy contradiction. (What are the factors of $p^2$?)
A: It's easy to see that $p^2+q^2=3^2$ has no solutions in primes (in fact there are no positive integers whose squares sum to $9$).  It's also easy to see that $3^2+q^3=r^2$ (and likewise $p^2+3^2=r^2$) has no solutions in primes:  $q$ cannot be odd, since $3^2+q^2$ would then be an even number greater than $4$, but it can't be $2$ either, since $3^2+2^2=13$ is not a square.  So we can conclude that $3\not\mid pqr$.  But in that case, $p^2\equiv q^2\equiv r^2\equiv1$ mod $3$, and we see that $p^2+q^2\equiv2\not\equiv1$ mod $3$.
A: If $p,q$ are both odd, then $r=2$, no solutions. If wlog $p=2$, then $4=(r-q)(r+q)$. But $r+q>4$, contradiction.
