I know that this may seem an easy/noob question. Sorry if you also think that.
As a math enthusiast, I have spent some hours looking for different proofs of the infinitude of prime numbers. Most of them (almost all) use the fact that every single integer has a unique prime factorization (Fundamental Theorem of Arithmetic) or, at least, that is a product of primes. However, I have not been able to find any proof that did not use that.
Then, this is what I am looking for. Let $f : \mathbb{N} \to \mathbb{N}$, such that $0 \le f(x) <x$. Also, let $p$ represent a prime number greater that $2$ and $m>0$ any arbitray integer.
Which ways are there to prove that there are infinitely many numbers that are not of the form $pm+f(p)$ for any prime $p>2$ without taking into account the fact that there are infinitely many primes?
I know that this "conjecture" would imply their infinitude. What I am asking for is a proof that does not contain the sentence "...and, as we already know that there are infinitely many prime numbers...". My problem when analyzing this problem is that we cannot "copy/paste" proofs of the infinitude of primes since here we cannot use the Fundamental Theorem of Arithmetic because of the addition.
Thank you.
Edit 1: we can make a basic heuristic to calculate the amount of these integer lower or equal than another given $x$.
$$ \# a \{a\not = pn+f(p); a \le x\} \sim x \prod_{p\le x} \frac{p-1}{p} =O \left ( \frac{x}{\log(x)} \right )$$
Edit 2: what is it known about the case $f(p)$ being a constant greater than $0$?