Prove that no points on a circle of radius $\sqrt{3}$ can have both $x$ and $y$ coordinates rational

Suppose, for the sake of contradiction, there was such an integer solution $p,m,n$, and so there must exist some coprime solution because if they share a common factor it can just cancel out. Then in particular $p^{2} + m^{2} \equiv 0$ modulo $3$, or in other words $p^{2} \equiv - m^{2}$ modulo $3$. Now since $-1$ is not a square modulo $3$, it follows that $p,m \equiv 0$ modulo $3$, and so we can write $p = 3p'$, $m = 3m'$ for some integers $p'$, $m'$, and then $n^{2} = 3({p'}^{2} + {m'}^{2})$ and so $n \equiv 0$ modulo $3$ which contradicts that $n,p,m$ are coprime (unless of course $p,n,m = 0$).