[Algebraic Number Theory I] Sum of Two Squares and Gaussian Integers

Reference(s):
[1] Neukirch, J. (1999). Algebraic number theory, volume 322 of Grundlehren der Mathematischen Wissenschaften [Fundamental Principles of Mathematical Sciences].
[2] Shoup, V. (2009). A computational introduction to number theory and algebra. Cambridge university press.



Hello everyone, I use this blog, owned by Thuong Dang, for practising my English writing skill in mathematics. This is my fist note and also the beginning of a series about algebraic number theory. The main reference I will use for this note, and this series, is from [1]. 

The main purpose of this note is to prove the following theorem:

Theorem 1 ([1, Theorem 1.1]). Every positive prime number $p$ in $\mathbb{N}$ is a sum of two squares, i.e., $p = a^2 + b^2$ for $a, b \in \mathbb{N}$, if and only if $p \equiv 1 \pmod{4}$.

To prove this theorem, we need to use the ring of Gaussian integers, i.e., the set
$$ \mathbb{Z}[i] = \{a + bi : a, b \in \mathbb{N}\} $$
equipped with the following two operations:
Addition. $(a + bi) + (c + di) = (a + c) + (b + d)i$, and
Multiplication. $(a + bi) * (c + di) = (ac - bd) + (cb + ad) i$.
For simplicity, we write $(a + bi)(c+di)$ in place of $(a + bi) * (c + di)$.

In this note, we only present the proof of Theorem 1. The underlying properties of the ring of Gaussian integers will be discussed in the next note.

We define the norm of a Gaussian integer $\alpha = a + bi$ to be following value:
$$ N(\alpha) = \alpha\overline{\alpha} = (a + bi)\overline{(a + bi)} = (a + bi)(a - bi) = a^2 + b^2.$$
Geometrically, the norm of a Gaussian integer $a + bi$ is the square of the distance from the point $(a,b)$ to the point $(0, 0)$ in a $2$-dimensional space.

Lemma 2 (Wilson's theorem, [2, Theorem 2.22]). Let $p$ be an odd prime. Then, we have
$$\prod_{z \in \mathbb{Z}^{*}_p}z \equiv -1 \pmod{p}.$$
Proof. We make pairs the elements in $\mathbb{Z}^{*}_p - \{-1, 1\}$ such that, for each pair (a, b), its product satisfies
$$a * b \equiv 1 \pmod{p}.$$
For a specific pair (a, b), we clearly see that $a \not\equiv b \pmod{p}$ because the only square roots of $1$ in modulo $p$ are $1$ and $-1$. Since $\mathbb{Z}_p^{*}$ forms a group structure, every element $a$ in $\mathbb{Z}^{*}_p - \{-1, 1\}$ has a unique partner $b \in \mathbb{Z}^{*}_p - \{-1, 1\}$ satisfying $a*b \equiv 1 \pmod{p}$. Therefore,
$$ \prod_{z \in \mathbb{Z} - \{-1, 1\}}z \equiv 1 \pmod{p} $$
and then,
$$ \prod_{z \in \mathbb{Z}}z \equiv -1 \pmod{p}$$
which proves the lemma.

Q.E.D

Lemma 3 ([1]). Let $p = 4n + 1$, $n \in \mathbb{N}$, be a prime. Then, $(2n)!$ is a solution of the modular equation $x^2 \equiv -1 \pmod{p}$.

Proof. Since $p$ is a prime, the elements of the set $\mathbb{Z} - \{0\}$ is exactly the elements of the group $\mathbb{Z}^{*}_p$. By Lemma 2, we have

$$\prod_{z = 1}^{p - 1}\equiv\prod_{z \in \mathbb{Z}^{*}_p}z \equiv -1 \pmod{p}. $$

Therefore, we have

$$ -1 \equiv 1 * 2 * ... * (2n) * [(2n + 1) * ... * 4n] $$ 
$$\equiv 1 * 2 * ... * (2n) * [(p - 1) * (p - 2)* ... * (p - 2n)] $$ 
$$\equiv 1*2*...*(2n)*[(-1) * (-2) *...* (-2n)]$$
$$\equiv 1*2* ...*(2n) * [1 * 2 * ... * (2n)] * (-1)^{2n}$$
$$\equiv [1*2*...*(2n)]^2 \pmod{p}.$$
Thus the lemma is proved.

Q.E.D

Now we prove the main result of this note.

Proof of Theorem 1. For the sufficient condition, since $p \equiv 1 \pmod{4}$, by Lemma 3, we have a solution $x$ of the equation 

$$ x^2 \equiv -1 \pmod{p}.$$

Hence, we have $x^2 + 1 \equiv 0 \pmod{p}$ or $p | (x^2 + 1).$ We see that 

$$x^2 + 1 = (x + i)(x-i). $$

Therefore, $p | (x + i)(x - i)$. Since $p$ is an odd prime, the fraction $\frac{1}{p} \not\in \mathbb{Z}$ and therefore, $p$ is a divisor of neither $x + i$ nor $x - i$. Hence, $x + i = (a + bi)(a'+b'i)$ and $x - i  = (c + di)(c'+d'i)$ such that the product $(a+bi)(c+di) = p$. We also note that the norm of the four Gaussian integers $a + bi, a'+b'i, c+di,$ and $c'+d'i$ are greater than $1$.

Consider $p^2 = N(p) = N(a + bi)N(c+di).$ Since $N(a+bi) > 1$ and $N(c+di) > 1$, we concludes that $N(a+bi) = p$ and $N(c+di) = p$. Thus, $p = N(a +bi) = a^2 + b^2$ which proves the sufficient condition.

For the necessary condition, consider an integer $z \in \mathbb{Z}$. We always have either $z^2 \equiv 0 \pmod{4}$ or $z^2 \equiv 1 \pmod{4}$. Hence, the value $a^2 + b^2$, for $a, b\in\mathbb{Z}$, cannot be equivalent to $3$ in modulo $4$. Thus the theorem is proved.

Q.E.D