Which Numbers are the Sum of Two Squares?

The main goal of today's lecture is to prove the following theorem.

Theorem 1.1   A number $n$ is a sum of two squares if and only if all prime factors of $n$ of the form $4m+3$ have even exponent in the prime factorization of $n$.

Before tackling a proof, we consider a few examples.

Example 1.2  

• $5 = 1^2 + 2^2$.
• $7$ is not a sum of two squares.
• $2001$ is divisible by $3$ because $2+1$ is, but not by $9$ since $2+1$ is not, so $2001$ is not a sum of two squares.
• $2\cdot 3^4\cdot 5\cdot 7^2\cdot 13$ is a sum of two squares.
• $389$ is a sum of two squares, since $389\equiv 1\pmod{4}$ and $389$ is prime.
• $21=3\cdot 7$ is not a sum of two squares even though $21\equiv 1\pmod{4}$.

In preparation for the proof of Theorem 1.1, we recall a result that emerged when we analyzed how partial convergents of a continued fraction converge.

Lemma 1.3   If $x\in\mathbb{R}$ and $n\in\mathbb{N}$, then there is a fraction $$\frac{a}{b}$$ in lowest terms such that $0<b\leq n$ and

$$\left\vert x - \frac{a}{b} \right\vert \leq \frac{1}{b(n+1)}.$$

Proof. Let $[a_0,a_1,\ldots]$ be the continued fraction expansion of $x$. As we saw in the proof of Theorem 2.3 in Lecture 18, for each $m$

$$\left\vert x - \frac{p_m}{q_m}\right\vert < \frac{1}{q_m \cdot q_{m+1}}.$$

Since $q_{m+1}$ is always at least $1$ bigger than $q_m$ and $q_0=1$, either there exists an $m$ such that $q_m\leq n < q_{m+1}$, or the continued fraction expansion of $x$ is finite and $n$ is larger than the denominator of the rational number $x$. In the first case,

$$\left\vert x - \frac{p_m}{q_m}\right\vert < \frac{1}{q_m \cdot q_{m+1}} \leq \frac{1}{q_m \cdot (n+1)},$$

so $$\frac{a}{b} = \frac{p_m}{q_m}$$ satisfies the conclusion of the lemma. In the second case, just let $$\frac{a}{b} = x$$.

$\qedsymbol$

Definition 1.4   A representation $n=x^2 + y^2$ is primitive if $\gcd(x,y)=1$.

Lemma 1.5   If $n$ is divisible by a prime $p$ of the form $4m+3$, then $n$ has no primitive representations.

Proof. If $n$ has a primitive representation, $n=x^2 + y^2$, then

$$p \mid x^2 + y^2$$    and $$\quad \gcd(x,y)=1,$$

so $p\nmid x$ and $p\nmid y$. Thus $x^2 + y^2 \equiv 0\pmod{p}$ so, since $\mathbb{Z}/p\mathbb{Z}$ is a field we can divide by $y^2$ and see that

$$(x/y)^2 \equiv -1\pmod{p}.$$

Thus the quadratic residue symbol $\left(\frac{-1}{p}\right)$ equals $+1$. However,

$$\left(\frac{-1}{p}\right) = (-1)^{\frac{p-1}{2}} = (-1)^\frac{4m+3-1}{2} = (-1)^{2m+1} = -1.$$

$\qedsymbol$

Proof. [Proof of Theorem 1.1] $\left(\Longrightarrow\right)$ Suppose that $p$ is of the form $4m+3$, that $p^r\mid\mid n$ (exactly divides) with $r$ odd, and that $n=x^2 + y^2$. Letting $d=\gcd(x,y)$, we have

$$x = dx', \quad y = dy', \quad n = d^2 n'$$

with $\gcd(x',y')=1$ and

$$(x')^2 + (y')^2 = n'.$$

Because $r$ is odd, $p\mid n'$, so Lemma 1.5 implies that $\gcd(x',y')>1$, a contradiction.

$\left(\Longleftarrow\right)$ Write $n=n_1^2 n_2$ where $n_2$ has no prime factors of the form $4m+3$. It suffices to show that $n_2$ is a sum of two squares. Also note that

$$(x_1^2 + y_1^2)(x_2^2+y_2^2) = (x_1x_2+y_1y_2)^2 + (x_1y_2-x_2y_1)^2,$$

so a product of two numbers that are sums of two squares is also a sum of two squares.¹Also, the prime $2$ is a sum of two squares. It thus suffices to show that if $p$ is a prime of the form $4m+1$, then $p$ is a sum of two squares.

Since

$$(-1)^{\frac{p-1}{2}} = (-1)^{\frac{4m+1-1}{2}} = +1,$$

$-1$ is a square modulo $p$; i.e., there exists $r$ such that $r^2\equiv -1\pmod{p}$. Taking $n=\lfloor \sqrt{p}\rfloor$ in Lemma 1.3 we see that there are integers $a, b$ such that $0<b<\sqrt{p}$ and

$$\left\vert -\frac{r}{p} - \frac{a}{b}\right\vert \leq\frac{1}{b(n+1)} < \frac{1}{b\sqrt{p}}.$$

If we write

$$c = rb + pa$$

then

$$\vert c\vert < \frac{pb}{b\sqrt{p}} = \frac{p}{\sqrt{p}} = \sqrt{p}$$

and

$$0 < b^2 + c^2 < 2p.$$

But $c \equiv rb\pmod{p}$, so

$$b^2 + c^2 \equiv b^2 + r^2 b^2 \equiv b^2(1+r^2) \equiv 0\pmod{p}.$$

Thus $b^2 + c^2 = p$. $\qedsymbol$