Sum of Four Squares¶
Theorem
Let \(p\) be a prime, for all \(a \in \mathbb{Z}\), the equation \(x^2 + y^2 \equiv a \pmod p\) has a solution.
Proof
If \(p = 2\), it's obvious.
If \(p\) is odd, then notice that there are \(\frac{p+1}{2}\) quadratic residues and \(\frac{p-1}{2}\) quadratic non-residues in \(\mathbb{Z}/p\mathbb{Z}\). The set of quadratic residues is called \(S\), \(\lvert S \rvert = \frac{p + 1}{2}\). Set \(S' = \{a - z \mid z \in S\}\), \(\lvert S' \rvert = \frac{p + 1}{2}\). Then \(\lvert S \rvert + \lvert S' \rvert = p + 1 > p\), which means \(S \cap S' \neq \varnothing\). So there exists \(x^2 \in S\), \(a - y^2 \in S'\), such that \(x^2 \equiv a - y^2 \pmod p\). So \(x^2 + y^2 \equiv a \pmod p\).
Theorem
(Lagrange) For any integer \(n\), the equation \(x^2+y^2+z^2+w^2 = n\) has a solution.
Proof
First prove that Lagrange is right for prime \(p\).
Definition
(Hamiltonian Quaternion) \(\mathbb{H} := \left\{\begin{pmatrix} a & b \\ -\bar{b} & \bar{a} \end{pmatrix} \mid a, b \in \mathbb{C} \right\}\)
So for \(x^2 + y^2 + z^2 + w^2 = n\), define \(\alpha = x+y\sqrt{-1}, \beta = z+w\sqrt{-1}\), \(x^2+y^2+z^2+w^2 = \alpha \bar{\alpha} + \beta \bar{\beta} = n\). Define \(\begin{pmatrix} \alpha & \beta \\ -\bar{\beta} & \bar{\alpha} \end{pmatrix} = A\). So we just solve the equation \(N(A) = n\) in \(\mathbb{H}.\)
Example
Express \(310\) as a sum of four squares.
We can do this by expressing \(10\) and \(31\) as a sum of four squares. We know that \(31 = 5^2 + 2^2 + 1^2 + 1^2\) and \(10 = 3^2 + 1^2 + 0^2 + 0^2\). So we can write two matrices.
So \(N(\alpha) = 31, N(\beta) = 10\), and \(N(\alpha \beta) = 310\). \(\alpha \cdot \beta = \begin{pmatrix} 13 + 11\mathrm{i} & 4 + 2\mathrm{i} \\ -4 + 2\mathrm{i} & 13 - 11\mathrm{i} \end{pmatrix}\). So \(310 = 13^2 + 11^2 + 4^2 + 2^2\). This is one of the solutions.