a. Show that 2 is equal to the product of a unit and the square of an irreducible in Z[i]. 

b. Show that an odd prime p in Z is irreducible in Z[i] if and only if p ≡ 3 (mod 4). (Use Theorem 47.10.)

Data from Theorem 47.10

Let p be an odd prime in Z. Then p = a² + b² for integers a and b in Z if and only if p ≡ 1 (mod 4).

Proof First, suppose that p = a² + b². Now a and b cannot both be even or both be odd since p is an odd number. If a= 2r and b = 2s + 1, then a² + b² = 4r² + 4(s² + s) + 1, so p ≡ 1 (mod 4). This takes care of one direction for this "if and only if" theorem. For the other direction, we assume that p ≡ 1 (mod 4). Now the multiplicative group of nonzero elements of the finite field Z_P is cyclic, and has order p - 1. Since 4 is a divisor of p - 1, we see that Z_P contains an element n of multiplicative order 4. It follows that n² has multiplicative order 2, so n² = -1 in Z_P. Thus in Z, we have n² ≡ -1 (mod p ), so p divides n² + 1 in Z.

Viewing p and n² + 1 in Z[i], we see that p divides n² + 1 = (n + i)(n - i). Suppose that p is irreducible in Z[i]; then p would have to divide n + i or n - i. If p divides n + i, then n + i = p(a + bi) for some a, b ∈ Z. Equating coefficients of i, we obtain 1 = pb, which is impossible. Similarly, p divides n - i would lead to an impossible equation -1 = pb. Thus our assumption that p is irreducible in Z[i] must be false. Since p is not irreducible in Z[i], we have p = (a + bi)(c + di) where neither a + bi nor c + di is a unit. Taking norms, we have p² = (a² + b²)(c² + d²) where neither a² + b² = 1 nor c² + d² = 1. Consequently, we have p = a² + b², which completes our proof. [Since a² + b² = (a + bi)(a - bi), we see that this is the factorization of p, that is, c + di = a - bi.]