# Proofs from the BOOK: Fermat's theorem on sums of two squares

I need some help in understanding that there are infinitely many prime numbers of the form $4$$m+3. (The author introduced this theorem as a warm-up) Let's assume that there are only finitely many prime numbers of form 4$$m$+$3$, with $p_k$ denoting the largest prime.

Express every prime number smaller than $p_k$ in order $p_1$ = 2, $p_2$ =3 ...

Suppose there is a natural number $N_k$

$N_k$ $=$ $p_1^2$$p_2$$p_3$ ... $p_k$ $-1$

Then $N_k$ $\equiv$ $3$ (mod $4$), so it must have prime factor of the form $4$$m+3 that is bigger than p_k. Can somebody explain how the fact that N_k \equiv 3 (mod 4) shows N_k must have a factor of the form 4$$m$+$3$ and why should it be bigger than $4$$m$+$3$?

Thank you in advance.

## 1 Answer

An odd number can have only odd factors. The product of two numbers of the form $4r+1$ is of the same form, so a number of the form $4n+3$ must have at least one factor of the form $4m+3$ - if all factors were of type $4r+1$, the product would be of form $4r+1$.

This is easy to see looking at things modulo $4$, when the odd numbers reduce to $\pm 1$ and you can't get the result $-1$ by multiplying $1$ by itself.

As for size $N_k=4\times 3 \times 5 \times p_k-1\ge 4p_k-1\gt p_k$

• Hi, thanks for the clarification! Now I can see things more clearly. – ecg Sep 23 '17 at 11:29