1. ## Gaussian Integers

In this part, you may assume any facts about the factorisation theory of Z[i], the ring of Gaussian Integers, and of Z provided that you state clearly the properties that you are using.
Let p be an integer prime for which there is an element a in Z with $a^2 + 1 = p$. Write down a factorisation of p in Z[i] and show that it is a proper factorisation (that is, neither factor is a unit of Z[i]). Deduce that p is not a prime in the Gaussian integers. Show that the factors of p that you obtained are primes in Z[i].

Thoughts...

p = (a + i)(a - i)
I'm not certain about this but i think that a prime cant be a unit? So in Z that would be $\pm 1$?
If so then $a \neq 0$ hence since the only units in Z[i] are 1,-1,i ,-i. Neither factor is not a unit, i.e. proper.

p is not prime since p doesn't divide either factor. This is because $a \pm i$ is imaginary and p is real.

Showing $a \pm i$ are primes in Z[i]...
a + bi is prime if $a^2 + b^2$ is an integer prime of the form 4k+1. So how do i show $a^2 + 1$ is of the form 4k+1?

Actually a thought has just occurred... Since $a^2 + 1$ is prime, $a^2$ has to be even, i.e. a is even. So let a be of the form 2n. Then we have $(2n)^2 + 1 = 4k + 1$ so $4n^2 = 4k$, hence $k = n^2$? So... ?

One final note... $a^2 + 1$ could be even, but only if a=1, since 2 is the only even prime. But 1+i and 1-i are primes in Z[i] so that takes care of the a=odd case.

In this part, you may assume any facts about the factorisation theory of Z[i], the ring of Gaussian Integers, and of Z provided that you state clearly the properties that you are using.
Let p be an integer prime for which there is an element a in Z with $a^2 + 1 = p$. Write down a factorisation of p in Z[i] and show that it is a proper factorisation (that is, neither factor is a unit of Z[i]). Deduce that p is not a prime in the Gaussian integers. Show that the factors of p that you obtained are primes in Z[i].

Thoughts...

p = (a + i)(a - i)
I'm not certain about this but i think that a prime cant be a unit? So in Z that would be $\pm 1$?
If so then $a \neq 0$ hence since the only units in Z[i] are 1,-1,i ,-i. Neither factor is not a unit, i.e. proper.

p is not prime since p doesn't divide either factor. This is because $a \pm i$ is imaginary and p is real.

Showing $a \pm i$ are primes in Z[i]...
a + bi is prime if $a^2 + b^2$ is an integer prime of the form 4k+1. So how do i show $a^2 + 1$ is of the form 4k+1?

Actually a thought has just occurred... Since $a^2 + 1$ is prime, $a^2$ has to be even, i.e. a is even. So let a be of the form 2n. Then we have $(2n)^2 + 1 = 4k + 1$ so $4n^2 = 4k$, hence $k = n^2$? So... ?

One final note... $a^2 + 1$ could be even, but only if a=1, since 2 is the only even prime. But 1+i and 1-i are primes in Z[i] so that takes care of the a=odd case.
Define $N: \mathbb{Z}[i]\to \mathbb{Z}[i]$ by $N(\alpha) = \alpha \bar \alpha$, so if $\alpha=a+bi$ then $N(\alpha) = a^2+b^2$.
It is easy to show that $\alpha$ is a unit if and only if $N(\alpha) = 1$.
To show that $a\pm 1$ is not a unit just notice that $N(a\pm 1) = (a\pm 1)^2 > 1$ (if $|a|> 1$).

If $N(\alpha) = p$ for a prime then $\alpha$ must be prime. Because otherwise $\alpha = \beta \gamma$ for non-units and so $p=N(\alpha) = N(\beta)N(\gamma)$. Which is a problem because $p$ is prime.