Exercise 1.3.12 (If $x$ and $y$ are prime to $3$, $x^2 + y^2$ cannot be a perfect square)

Proof. Assume for contradiction that $x$ and $y$ are prime to $3$, and that $x^2+y^2=z^2$ is a perfect square. Then $x\equiv \pm 1(\mathit{mod}3)$, and $y\equiv \pm 1(\mathit{mod}3)$, thus $x^2\equiv y^2\equiv 1(\mathit{mod}3)$. Therefore

But this is impossible, because the square of any integer is of the form $3k$ or $3k+1$, but not of the form $3k+2$ (see Problem 1.2.22). □