Exercise 3.6.5 (Number of ordered pairs $(x,y)$ of positive integers for which $x^2 + y^2 = n$)

Proof.

(a)

Let $n$ be a positive integer, and let $S$ be the set of solutions $(x,y)\in {\mathbb{Z}}^2$ of $x^2+y^2=n$. Put $$\begin{array}{llll}\hfill S_1& =\{(x,y)\in S_n\mid x>0,y\ge 0\},& \hfill & \\ \hfill S_2& =\{(x,y)\in S_n\mid y>0,x\ge 0\},& \hfill & \\ \hfill S_3& =\{(x,y)\in S_n\mid x<0,y\le 0\},& \hfill & \\ \hfill S_4& =\{(x,y)\in S_n\mid y>0,x\ge 0\},& \hfill & \\ \hfill & & \hfill \end{array}$$

Since $(0,0)$ is not a solution,

$$S=S_1\cup S_2\cup S_3\cup S_4,$$

and this union is a disjoint union, therefore

$$R(n)=|S|=|S_1|+|S_2|+|S_3|+|S_4|.$$

Moreover, the maps

$$\phi \{\begin{array}{ccc}\hfill S_1\hfill & \hfill \to \hfill & S_2\hfill \\ \hfill (x,y)\hfill & \hfill \mapsto \hfill & (-y,x)\hfill \end{array}\psi \{\begin{array}{ccc}\hfill S_2\hfill & \hfill \to \hfill & S_1\hfill \\ \hfill (x,y)\hfill & \hfill \mapsto \hfill & (y,-x)\hfill \end{array}$$

(rotations of angles $+\pi ∕2$, $-\pi ∕2$, restricted to $S_1$ and $S_2$) satisfy $\psi \circ \phi =1_{S_1},\phi \circ \psi =1_{S_2}$, thus $\phi$ is a bijection, so $|S_1|=|S_2|$, and similarly $|S_2|=|S_3|=|S_4|$. Therefore

$$R(n)=4|S_1|,$$

so

$$\begin{array}{lll}\hfill R(n)=4\mathrm{Card}\{(x,y)\in {\mathbb{Z}}^2\mid x^2+y^2=n,x>0,y\ge 0\}.& & \hfill \text{(1)}\end{array}$$

Suppose that $n$ is not a perfect square. Then there is no solution of $x^2+y^2=n$ such that $x=0$ or $y=0$, thus

$$S_1=\{(x,y)\in {\mathbb{Z}}^2\mid x^2+y^2=n,x>0,y>0\}.$$

Then the equality (1) shows that

$$\mathrm{Card}\{(x,y)\in {\mathbb{Z}}^2\mid x^2+y^2=n,x>0,y>0\}=R(n)∕4.$$

The number of ordered pairs $(x,y)$ of positive integers for which $x^2+y^2=n$ is $R(n)∕4$.

(b)

Suppose now that $n=a^2$ is a perfect square. Then, with the notations of part (a), $$\begin{array}{llll}\hfill S_1& =\{(x,y)\in {\mathbb{Z}}^2\mid x^2+y^2=n,x>0,y\ge 0\}& \hfill & \\ \hfill & =\{(a,0)\}\cup \{(x,y)\in {\mathbb{Z}}^2\mid x^2+y^2=n,x>0,y>0\}.& \hfill & \end{array}$$

Therefore

$$R(n)∕4=|S_1|=1+\mathrm{Card}\{(x,y)\in {\mathbb{Z}}^2\mid x^2+y^2=n,x>0,y>0\},$$

so

$$\mathrm{Card}\{(x,y)\in {\mathbb{Z}}^2\mid x^2+y^2=n,x>0,y>0\}=R(n)∕4-1.$$

If $n$ is a perfect square then the number of ordered pairs $(x,y)$ of positive integers for which $x^2+y^2=n$ is $R(n)∕4-1$