Exercise 3.3.8 (Congruence $x^2 + y^2 \pmod {p^\alpha}$)

Proof. Suppose that $x^2+y^2\equiv 0(mod{p}^{\alpha })(\alpha \ge 1)$, where $x\wedge p=y\wedge p=1$. A fortiori $x^2+y^2\equiv 0(modp)$. By Problem 8,

Conversely, suppose that $p=2\text{ or }p\equiv 1(mod4)$. By Problem 8, there are some integers $x_1,y_1$ such that $x_1^2+y_1^2\equiv 0(modp)$, where $x_1\wedge p=y_1\wedge p=1$.

• Suppose first that $p\ne 2$.

  Put $f(x,y)=x^2+y^2$. Here $f$ is a polynomial with integral coefficients in the two variables $x,y$, and

  $$\begin{array}{llll}\hfill \frac{\partial f}{\partial x}(x_1,y_1)& =2x_1,& \hfill & \\ \hfill \frac{\partial f}{\partial y}(x_1,y_1)& =2y_1.& \hfill & \end{array}$$

  Since $p$ is an odd prime, and $p\nmid x_1$, $\frac{\partial f}{\partial x}(x_1,y_1)\not\equiv 0(modp)$. By the generalization of Hensel’s Lemma given in Problem 2.6.11, the congruence $x^2+y^2\equiv 0(mod{p}^{\alpha })$ has a solution, for every $\alpha$.

• Consider now that case $p=2$, where $\left(\frac{\partial f}{\partial x}(x_1,y_1),\frac{\partial f}{\partial y}(x_1,y_1)\right)=(0,0)$. The congruence $x^2+y^2\equiv 0(mod2)$ has a solution $(x_1,y_1)=(1,1)$. But the congruence $x^2+y^2\equiv 0(mod4)$ has no solution $(x,y)$ where $x,y$ are odd integers: if $x,y$ are odd, then $x^2\equiv y^2\equiv 1(mod4)$, thus $x^2+y^2\equiv 2\not\equiv 0(mod4)$. A fortiori $x^2+y^2\equiv 0(mod{2}^{\alpha })$ has no solution $(x,y)\in {\mathbb{Z}}^2$ satisfying $x\wedge 2=y\wedge 2=1$ for $\alpha \ge 2$.

To conclude, there exist integers $x$ and $y$ with $(x,p)=1,(y,p)=1$, such that $x^2+y^2\equiv 0(mod{p}^{\alpha })$ if and only if