Sum of Two Squares¶
Primes in \(\mathbb{Z}[\sqrt{-1}]\)¶
Main Theorem
(The primes in \(\mathbb{Z}[\sqrt{-1}]\))
(i) 2 is a prime in \(\mathbb{Z}\), \(2 = (1+\sqrt{-1})^2 \cdot (-\sqrt{-1})\).
(ii) If \(p \equiv 3 \pmod 4\) is a prime in \(\mathbb{Z}\), then \(p\) stays as a prime in \(\mathbb{Z}[\sqrt{-1}]\).
(iii) If \(p \equiv 1 \pmod 4\) is a prime in \(\mathbb{Z}\), then \(p = \mathcal{p} \cdot \bar{\mathcal{p}}\), \(\mathcal{p}\) and \(\bar{\mathcal{p}}\) are primes in \(\mathbb{Z}[\sqrt{-1}]\).
(iv) All primes in \(\mathbb{Z}[\sqrt{-1}]\) are listed above.
Lemma
(i) If \(p\) is a prime in \(\mathbb{Z}[\sqrt{-1}]\), but \(p \not \in \mathbb{Z}, p \not \in i\mathbb{Z}\). Then \(N(p)\) is a prime in \(\mathbb{Z}\), and \(N(p) \equiv 1 \pmod 4\) or \(N(p) = 2\).
(ii) If \(p \neq 2, p \in \mathbb{Z}\). Then \(p\) is a prime \(\mathbb{Z}[\sqrt{-1}]\) iff \(p \equiv 3 \pmod 4\).
Proof
(i) \(N(p) = p \cdot \bar{p} \in \mathbb{Z}\). According to the factorization in \(\mathbb{Z}\), we have \(N(p) = p_1^{r_1}p_2^{r_2}\cdots p_k^{r_k}\). So \(p \mid p_1^{r_1}p_2^{r_2}\cdots p_k^{r_k}\) in \(\mathbb{Z}[\sqrt{-1}]\).
Without loss of generality, we assume \(p \mid p_1\), then we have \(p_1 = p \cdot \alpha, \alpha \in \mathbb{Z}[\sqrt{-1}]\). \(p_1^2 = N(p_1) = N(p) \cdot N(\alpha)\), \(N(p) \neq p_1^2\) because \(p \not \in \mathbb{Z}, p \not \in i\mathbb{Z}\). So \(N(p) = p_1\). \(\alpha = \dfrac{p_1}{p} = \dfrac{p_1 \cdot \bar{p}}{p \cdot \bar{p}} = \bar{p}\), which means \(p_1 = p \cdot \bar{p}\), and then \(N(p) = p \cdot \bar{p} = p_1\).
Let \(p = x + y\sqrt{-1}\). If \(N(p) = 2\), it's true. If \(N(p) \neq 2\), then \(p_1 = x^2 + y^2\), \(p_1 = N(p) \equiv 1 \pmod 4\).(See in Lec3)
(ii) If \(p \equiv 1 \pmod 4\), then \(p = x^2 + y^2\) has a solution in \(\mathbb{Z} \times \mathbb{Z}\), and \(p = (x+y\sqrt{-1})(x-y\sqrt{-1})\) is not a prime.
It remains to prove that \(p \equiv 3 \pmod 4 \Rightarrow p\) is a prime in \(\mathbb{Z}[\sqrt{-1}]\). If \(p\) were not a prime, then \(p = \alpha \cdot \beta\) for some \(\alpha, \beta \not \in (\mathbb{Z}[\sqrt{-1}])^{\times}\). Then \(p^2 = N(p) = N(\alpha)N(\beta)\), \(N(\alpha) = p\), \(\beta = \bar{\alpha}\), \(p = x^2 + y^2\). Contradiction!
Sum of Two Squares¶
Main Theorem
\(n = p_1^{r_1}\cdots p_k^{r_k}\)(unique factorization in \(\mathbb{Z}\)). Then \(x^2+y^2 = n\) has a solution iff \(r_j\) is even whenever \(p_j \equiv 3 \pmod 4\).
Lemma
If \(x_1^2 + y_1^2 = n_1, x_2^2 + y_2^2 = n_2\), then \((x_1x_2 + y_1y_2)^2+(x_1y_2 - x_2y_1)^2 = n_1n_2\).
Proof of Main Theorem
\((\Leftarrow)\) Suppose \(n = p_1^{r_1}\cdots p_k^{r_k} \cdot q_1^{s_1}\cdots q_l^{s_l}\), where \(p_j \equiv 1 \pmod 4, q_j \equiv 3 \pmod 4\) and \(s_j\) is even. Because \(p_j \equiv 1 \pmod 4\), \(p_j = x_j^2 + y_j^2\) has a solution. And \(q_j^{s_j} \equiv 1 \pmod 4\), so \(x^2+y^2 = q_j^{s_j}\) has a solution. Then according to the lemma, \(x^2+y^2 = n\) has a solution.
\((\Rightarrow)\) If \(x^2 + y^2 = n\) has a solution \((x_0, y_0)\), \(p_j \equiv 3 \pmod 4\). We need to prove \(r_j\) is even. \(p_j \mid n = (x_0+y_0\sqrt{-1})(x_0-y_0\sqrt{-1})\). Recall that \(p_j \equiv 3 \pmod 4 \Rightarrow p_j\) is a prime in \(\mathbb{Z}[\sqrt{-1}]\). \(p_j \mid x_0+y_0\sqrt{-1} \Rightarrow p_j \mid x_0, p_j \mid y_0\). Then \(p_j^2 \mid n\), which means \(x^2+y^2 = \dfrac{n}{p_j^2}\) has a solution \((\dfrac{x_0}{p_j}, \dfrac{y_0}{p_j})\).
Prood by induction. Induction hypothesis: For all integer \(k < n\), if \(x^2+y^2 = k\) has a solution, then \(r_j \equiv 0 \pmod 2\) whenever \(p_j \equiv 3 \pmod 4\). By induction hypothesis, the exponent of \(p_j\) in the factorization of \(\dfrac{n}{p_j^2}\) is even. Then the exponent of \(p_j\) in the factorization of \(n\) is even.
The Number of Solutions and Dirichlet \(L\)-Function¶
Lemma
\(x^2+y^2 = n\) has a solution \(\alpha = x_0+y_0\sqrt{-1}\). Then in \(\alpha, -\alpha, \sqrt{-1}\alpha, -\sqrt{-1}\alpha\), there exists a unique solution such that real part \(> 0\), imaginary part \(\geqslant 0\).