Let a Pythagorean triple $(\xi,n,\zeta)$.Prove that at least one of $\xi,n$ is divided by $3$ and that at least one of $\xi,n, \zeta$ is divided by $5$.
That's what I have thought:
We suppose that : 1)$\xi=3n+1 \text{ and } n=3m+1$
2)$\xi=3n+1 \text{ and }n=3m+2$
3) $\xi=3n+2 \text{ and } n=3m+2$
and we show that at no of the above cases it is possible that $\xi^2+n^2=\zeta^2\text{ ,because the remainders will be } 2,8,9, \text{ but the only possible remainders that can give } n^2,\text{ divided by } 9, \text{ are } 0,1, \text{ so it is a contradicton.} $
Similarly,we can show that at least one of $\xi,n, \zeta$ must be divisible by $5$.
Is the way I thought right??
Is there also an other way?
The only squares modulo three are $0$ and $1$, if neither $\xi$ or $n$ are divisible by $3$ we have $\zeta^2 \equiv 2 \pmod 3$, impossible.
The only squares modulo $5$ are $0,1,-1$ so if none of $\xi,n \zeta$ are divisible by $5$, we have $1+ 1\equiv \pm \zeta^2 \pmod 5$ or $1- 1\equiv \pm \zeta^2 \pmod 5$ also seen to be impossible.