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  

• <!---->.
• 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.
• <!----> is a sum of two squares.
• is a sum of two squares, since <!----> and is prime.
• <!----> is not a sum of two squares even though <!---->.

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 <!----> and <!----> , then there is a fraction <!----> in lowest terms such that and <!---->

Proof. Let <!----> be the continued fraction expansion of  . As we saw in the proof of Theorem 2.3 in Lecture 18, for each  <!---->

Since is always at least  bigger than and , either there exists an  such that <!----> , or the continued fraction expansion of  is finite and is larger than the denominator of the rational number  . In the first case, <!---->

so <!----> satisfies the conclusion of the lemma. In the second case, just let <!----> .

Definition 1.4   A representation <!----> is primitive if <!----> .

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

Proof. If  has a primitive representation, <!----> , then <!---->

    and

so and . Thus <!----> so, since <!----> is a field we can divide by and see that <!---->

Thus the quadratic residue symbol <!----> equals . However, <!---->

Proof. [Proof of Theorem  1.1] <!----> Suppose that  is of the form , that <!----> (exactly divides) with  odd, and that . Letting <!----> , we have <!---->

with <!----> and <!---->

Because  is odd, , so Lemma 1.5 implies that <!---->, a contradiction.

<!----> Write <!----> where has no prime factors of the form . It suffices to show that  is a sum of two squares. Also note that <!---->

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 <!---->

is a square modulo  ; i.e., there exists  such that <!----> . Taking <!----> in Lemma  1.3 we see that there are integers such that <!----> and <!---->

If we write <!---->

then <!---->

and <!---->

But <!----> , so <!---->

Thus <!----> .