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 is a sum of two squares if and only if all prime factors of of the form have even exponent in the prime factorization of.

Before tackling a proof, we consider a few examples.

Example 1.2

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

In preparation for the proof of Theorem1.1, we recall a result that emerged when we analyzed how partial convergents of a continued fraction converge.

Lemma 1.3 If <!-- MATH $x\in\mathbb{R}$ --> and <!-- MATH $n\in\mathbb{N}$ -->, then there is a fraction <!-- MATH $\displaystyle \frac{a}{b}$ --> in lowest terms such that and <!-- MATH \begin{displaymath} \left| x - \frac{a}{b} \right| \leq \frac{1}{b(n+1)}. \end{displaymath} -->

Proof. Let <!-- MATH $[a_0,a_1,\ldots]$ --> be the continued fraction expansion of. As we saw in the proof of Theorem2.3 in Lecture18, for each<!-- MATH \begin{displaymath} \left| x - \frac{p_m}{q_m}\right| < \frac{1}{q_m \cdot q_{m+1}}. \end{displaymath} -->

Since is always at least bigger than and , either there exists an such that <!-- MATH $q_m\leq n < q_{m+1}$ -->, or the continued fraction expansion of is finite and is larger than the denominator of the rational number. In the first case, <!-- MATH \begin{displaymath} \left| x - \frac{p_m}{q_m}\right| < \frac{1}{q_m \cdot q_{m+1}} \leq \frac{1}{q_m \cdot (n+1)}, \end{displaymath} -->

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

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

Lemma 1.5 If is divisible by a prime of the form , then has no primitive representations.

Proof. If has a primitive representation, <!-- MATH $n=x^2 + y^2$ -->, then <!-- MATH \begin{displaymath} p \mid x^2 + y^2\quad \text{ and }\quad \gcd(x,y)=1, \end{displaymath} -->

and

so and . Thus <!-- MATH $x^2 + y^2 \equiv 0\pmod{p}$ --> so, since <!-- MATH $\mathbb{Z}/p\mathbb{Z}$ --> is a field we can divide by and see that <!-- MATH \begin{displaymath} (x/y)^2 \equiv -1\pmod{p}. \end{displaymath} -->

Thus the quadratic residue symbol <!-- MATH $\left(\frac{-1}{p}\right)$ --> equals . However, <!-- MATH \begin{displaymath} \left(\frac{-1}{p}\right) = (-1)^{\frac{p-1}{2}} = (-1)^\frac{4m+3-1}{2} = (-1)^{2m+1} = -1. \end{displaymath} -->

Proof. [Proof of Theorem1.1] <!-- MATH $\left(\Longrightarrow\right)$ --> Suppose that is of the form , that <!-- MATH $p^r\mid\mid n$ --> (exactly divides) with odd, and that . Letting <!-- MATH $d=\gcd(x,y)$ -->, we have <!-- MATH \begin{displaymath} x = dx', \quad y = dy', \quad n = d^2 n' \end{displaymath} -->

with <!-- MATH $\gcd(x',y')=1$ --> and <!-- MATH \begin{displaymath} (x')^2 + (y')^2 = n'. \end{displaymath} -->

Because is odd, , so Lemma1.5 implies that <!-- MATH $\gcd(x',y')>1$ -->, a contradiction.

<!-- MATH $\left(\Longleftarrow\right)$ --> Write <!-- MATH $n=n_1^2 n_2$ --> where has no prime factors of the form . It suffices to show that is a sum of two squares. Also note that <!-- MATH \begin{displaymath} (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, \end{displaymath} -->

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

Since <!-- MATH \begin{displaymath} (-1)^{\frac{p-1}{2}} = (-1)^{\frac{4m+1-1}{2}} = +1, \end{displaymath} -->

is a square modulo; i.e., there exists such that <!-- MATH $r^2\equiv -1\pmod{p}$ -->. Taking <!-- MATH $n=\lfloor \sqrt{p}\rfloor$ --> in Lemma1.3 we see that there are integers such that <!-- MATH $0<b<\sqrt{p}$ --> and <!-- MATH \begin{displaymath} \left| -\frac{r}{p} - \frac{a}{b}\right| \leq\frac{1}{b(n+1)} < \frac{1}{b\sqrt{p}}. \end{displaymath} -->

If we write <!-- MATH \begin{displaymath} c = rb + pa \end{displaymath} -->

then <!-- MATH \begin{displaymath} |c| < \frac{pb}{b\sqrt{p}} = \frac{p}{\sqrt{p}} = \sqrt{p} \end{displaymath} -->

and <!-- MATH \begin{displaymath} 0 < b^2 + c^2 < 2p. \end{displaymath} -->

But <!-- MATH $c \equiv rb\pmod{p}$ -->, so <!-- MATH \begin{displaymath} b^2 + c^2 \equiv b^2 + r^2 b^2 \equiv b^2(1+r^2) \equiv 0\pmod{p}. \end{displaymath} -->

Thus <!-- MATH $b^2 + c^2 = p$ -->.