0

I have some questions after reading this post: Fermat's Christmas theorem on sums of two squares with Gaussian integers.

In this post, the op provides a proof of Fermat's theorem on sums of two squares. However, I'm confused about the following part:

It said that $p$ is not a Gaussian prime. Thus $$\begin{align*} p=\alpha \beta=(a+bi)(c+di) \end{align*}$$ I think this means that $p$ can be factored into the multiplication of two complex numbers.

My questions are

  1. Are $a+bi$ and $c+di$ Gaussian primes?
  2. Why are the norms of two complex numbers $\alpha$ and $\beta$ greater than one?
    Any help on this? Thanks.
M_k
  • 2,005
  • 5
  • 16
  • 2
    You do not assume they are primes, just that they are Gaussians that are not units, And because they are not units, their complex norm is not $1$. The complex norm of $a+bi$ is $1$, if and only if $(a+bi)(a-bi)=1$, if and only if $a+bi$ has multiplicative inverse $a-bi$. – Arturo Magidin Feb 11 '23 at 02:05
  • @Arturo Magidin Then I want to ask why none of them is a unit. I know units in Gaussian integers are only $\pm 1$ and $\pm i$. If, for example, I let $c+di=1$, then $p=a+bi$, and since $p$ is real, I can get $b=0$. I'm not really able to see what makes $a+bi$ and $c+di$ not units here. – M_k Feb 11 '23 at 02:12
  • If $p$ is not a prime, then it can be factored as a product of two Gaussians that are non-units, so you take two non-units whose product is $p$. You choose them to be non-units, witnesses to the asserted fact that $p$ is not a prime in the Gaussians. – Arturo Magidin Feb 11 '23 at 02:52
  • Thanks. About p is not a Gaussian prime, then it can be factored into the product of two non-units. Is there a proof of it? – M_k Feb 11 '23 at 02:56
  • 2
    Primes are always irreducible elements, and the definition of irreducible is that it is a non-unit such that if you factor it as a product of two factors, then one of the factors is a unit. (In the Gaussians, you also have Unique Factorization, so irreducibles are primes, but that does not hold in arbitrary integral domains). – Arturo Magidin Feb 11 '23 at 03:00

0 Answers0