## To show that alpha = epsilon*gamma^2 for some unit epsilon and some quadratic integers gamma in Q[i], we need to use the fundamental theorem of arithmetic in the Gaussian integers, Q[i].

Given that 32 = alpha*beta, where alpha and beta are relatively prime quadratic integers in Q[i], we want to express alpha in the form of epsilon*gamma^2.

Step 1: Find the prime factorization of 32 in Q[i].

First, factorize 32 in the Gaussian integers. We can write 32 as (2 + 2i)(2 - 2i)(2 + i)(2 - i). These are the prime factorization of 32 in Q[i].

Step 2: Write alpha and beta using their prime factorization.

Let's write alpha as (a + bi)(a - bi)(c + di)(c - di), where a, b, c, and d are integers.

Similarly, beta can be written as (x + yi)(x - yi)(z + wi)(z - wi), where x, y, z, and w are integers.

Step 3: Combine the positive and negative prime factors.

Since alpha and beta are relatively prime, their prime factors cannot be the same. Therefore, we can see that the prime factors in (a + bi)(a - bi) and (x + yi)(x - yi) cannot be the same.

Let's consider the prime factors (2 + 2i)(2 - 2i) for alpha.

(a + bi)(a - bi) must have either (2 + 2i) or (2 - 2i) as one of the prime factors, but not both.

Let's assume (2 + 2i) is a factor, so (2 - 2i) is not. We can express (2 + 2i) = (2 + i)*(1 + i).

Now, let's rewrite alpha = (a + bi)(a - bi)(c + di)(c - di) with these new factors:

alpha = (2 + 2i)(1 + i)(a + bi)(c + di)

Step 4: Group the prime factors.

Let's define epsilon = (2 + 2i)(1 + i), which is a unit in Q[i] because its norm |epsilon|^2 = |(2 + 2i)(1 + i)|^2 = |6 + 2i|^2 = 36 + 4 = 40 > 1.

Now, let's rewrite alpha using the grouping of prime factors:

alpha = epsilon(a + bi)(c + di)

Step 5: Define gamma.

We can define gamma = (a + bi)(c + di), where gamma is a quadratic integer in Q[i].

Finally, we can write alpha = epsilon*gamma^2, as desired.

Therefore, we have shown that alpha = epsilon*gamma^2 for some unit epsilon and some quadratic integer gamma in Q[i].