Number Fields - Sum of Two Squares

Sum of Two Squares

Theorem: Every prime $p = 1 (\mod 4)$ is a sum of two squares.

Proof: Let $p = 4 m + 1$ . By Wilson's Theorem, $n = (2 m)!$ is a square root of -1 modulo $p$ . (Alternatively, if $g$ is a primitive root of $ℤ_{p}^{*}$ we may take $n = g^{m}$ .)

Thus $p ∣ n^{2} + 1 = (n + i) (n - i)$ . Therefore $p$ is not irreducible, because it does not divide either of these factors.

Or is it? We overlooked a detail: how do we know primes and irreducibles are the same thing in $ℤ [i]$ ?

We answer this by showing $ℤ [i]$ is euclidean with respect to the norm $N (x + y i) = x^{2} + y^{2}$ . That is, for any $a, b \in ℤ [i]$ and $b$ nonzero, we find $q, r \in ℤ [i]$ with $a = b q + r$ and $N (r) < N (b)$ .

We find $q$ with $N (a / b - q) < 1$ via a geometric argument. The gaussian integers form a lattice, and $a / b$ lies within norm 1 of at least one of the points on this lattice, and we can take any of them to be $q$ .

Thus $ℤ [i]$ is euclidean and hence a UFD, so all irreducibles are prime.

Now we're sure $p$ is reducible, namely $p = (a + b i) (c + d i)$ for nonunits $a + b i, c + d i$ . Taking norms gives $p^{2} = (a^{2} + b^{2}) (c^{2} + d^{2})$ . Since neither factor of $p$ has norm 1, we conclude that $p = a^{2} + b^{2} = c^{2} + d^{2} ▪$

Armed with this result, we can now easily describe the units and primes of $ℤ [i]$ . The units are simply $\pm 1, \pm i$ , by considering elements of norm 1. As for the primes, if $p = 3 (\mod 4)$ is a prime in $ℤ [i]$ then it is also prime in $ℤ [i]$ since in this case $p$ cannot be the sum of two squares. Also if $a^{2} + b^{2}$ is prime, then $a + b i$ is also a prime.

We claim there are no other primes. To see this, let $π = a + b i$ be some prime in $ℤ [i]$ . Then $N (π) = p_{1} p_{2} . . . p_{k}$ for some primes $p_{i}$ in $ℤ$ . This implies $π ∣ p$ for some prime $p$ in $ℤ$ , and hence $N (π) = p$ or $N (π) = p^{2}$ . In the first case we have $a^{2} + b^{2} = p$ , and in the second, we see that $p / π$ is an integer of norm 1 implying that $π$ is an associate of $p$ . In the latter case, we must have $p = 3 (\mod 4)$ otherwise it would contradict what we proved above.

We can also show that a positive integer is the sum of two squares if and only if it has the form $a^{2} b$ where $b$ is squarefree and no prime factors equal to 3 modulo 4. This follows from the algebraic identity $(a^{2} + b^{2}) (c^{2} + d^{2}) = (a d - b c)^{2} + (a c + b d)^{2}$ , and the fact that the existence of nontrivial solutions of $x^{2} + y^{2} = 0 \mod p$ implies we can find a square root of -1 modulo $p$ , which is impossible if $p = 3 \mod 4$ .