Showing that an element is prime in $\mathbb{Z}$[i]

Question

Let p be a prime integer, and suppose p = a² + b² has NO integer solution.

The exercise asks that if p = a² + b² has no solution, then p is a prime in the set of Gaussian integers $\mathbb{Z}$[i], where if α $\in \mathbb{Z}$[i], then α = a + bi, where a, b are integers.

I solved this already, but I am reading a much shorter solution: If p = αβ, then p² = |α|² |β|². If |α|= 1, then α is a unit, and similarly for β. If we want p to not be a prime, then:
|α|² = |β|² = p, which implies p = |α|² = a² + b², which has no solution and thus p is a prime in the set of Gaussian integers.

My questions:
I'm having trouble understanding the second to last point: If α, β are not units, then why exactly does |α|² = |β|² = p?
I'm also having trouble understanding why p² = |α|² |β|². I'm not sure if it's a typo in the solutions, but from my understanding I thought that we can only write |p|² = |α|² |β|². What let's us write p² = |α|² |β|²?

Answer 1

Well, we said if $|\alpha| = 1$, then $\alpha$ is a unit, right? So that means if $\alpha$ is not a unit, then $|\alpha| \neq 1$.

So, if $p^{2} = |\alpha|^{2}|\beta|^{2}$, and both $\alpha, \beta$ are not units, then $|\alpha| \neq 1$ and $|\beta| \neq 1$, which means $|\alpha|^{2} \neq 1$ and $|\beta|^{2} \neq 1$.

But we know $$p^{2} = |\alpha|^{2} |\beta|^{2}$$

which implies $$ \underbrace{p p}_{\text{prime factorization of $p^{2}$}} = |\alpha|^{2} |\beta|^{2}$$

If both factors on the right hand side are not $1$, and $p$ is prime, then it should be clear (if it's not clear, please comment so I can clarify) that $|\alpha|^{2} = |\beta|^{2} = p$. (Remember that $p$ is a prime integer, and $|\alpha|, |\beta|$ are both positive integers.)

Answer 2

If the product of two natural numbers (both not $1$) is the square of a prime $p$, they obviously are both equal to $p$ by unique factorization.