I found this execise in Beachy and Blair: Abstract algebra:
Let $a,b,c$ be a Pithagorean triple. Show that either $a$ or $b$ is divisible by $3$ and one of $a,b,c$ is divisible by $5$.
Here is my try:
Assume that $a,b,c\in\mathbb{Z}$ and that $a^2+b^2=c^2$. If we look at the residues of the squares modulo $3$ we see that it is either $0$ or $1$ and here we arrive at the first result right away since if both $a$ and $b$ gives nonzero residue so $c$ gets an impossible residue modulo $3$. The same technique can be used for the second part but this time reducing modulo $5$. The possible residues are $\{0,1,4\}$. Once again we look at $a$ and $b$, if they give equal nonzero residue then we get an impossible residue for $c$ and if they give different nonzero residue then $c$ gives zero residue and hence the result follows.
Is this reasoning correct?
Is there any other way arriving at any of these results?
Thank you!
Yes, your reasoning is fine. :)