Exercise 5.3.2 (At least one of $x,y$ is divisible by $3$; at least one of $x,y,z$ is divisible by $5$)

Proof. Let $(x,y,z)\in {\mathbb{Z}}^3$ be a Pythagorean triple, so that $x^2+y^2=z^2$.

(a)

Assume for the sake of contradiction that $x\not\equiv 0,y\not\equiv 0(\mathit{mod}3)$. Then $$x^2\equiv y^2\equiv 1(\mathit{mod}3),$$

thus $z^2\equiv 2(\mathit{mod}3)$. But this is impossible, since $2$ is not a square modulo $3$. This contradiction shows that $3\mid x$ or $3\mid y$.

(b)

Suppose that $x\not\equiv 0,y\not\equiv 0(\mathit{mod}5)$. Then $$x^2\equiv \pm 1(\mathit{mod}5),y^2\equiv \pm 1(mod5),$$

thus $z^2=x^2+y^2\equiv 0,2,-2(\mathit{mod}5)$. But $2,-2$ are not squares modulo $5$, therefore $z^2\equiv 0(\mathit{mod}5)$, so $z\equiv 0(\mathit{mod}5)$. This shows that at least one of $x,y,z$ is divisible by $5$.