Apparently there is a way that I can use the fact that for any positive integer $n$ there are positive integers $a$ and $b$ such that $n=a^2b$ where $b$ is the possibly empty product of distinct primes (i.e squarefree?) to show that there are infinitely many primes.
My current tactic (the one that was hinted to me) was to fix x then assuming there are a finite number of primes show that there are only a finite number of possibilities for $n=a^2b$ knowing that $a^2 \leq b$
I believe it is mostly that last part that causes me issues as I am not sure how we can only consider $a^2$ and not $b$. I also assume that I have to do something with divisibility here though I am not sure what. I can't help but think that the overall approach is reminiscent of Euclid's though I am not sure?
Best Answer
If you know that every positive natural number can be written as $a^2 b$ with $b$ being the product of distinct primes (not dividing $a$), just assume that there are just $M$ primes $p_1,p_2,\ldots,p_M$ and consider $$ \sum_{n=1}^{N}\frac{1}{n} \tag{1}$$ for some large $N$. It is well known that such sum behaves like $\log(N)$. On the other hand, every $n\in[1,N]$ can be written as $a^2 b$ with $a\leq \sqrt{N}$ and $b$ being the product of the elements of a (possibly empty) subset of $\{p_1,\ldots,p_M\}$. In particular:
$$ \sum_{n=1}^{N}\frac{1}{n} \leq \sum_{a\leq \sqrt{N}}\frac{1}{a^2} \prod_{k=1}^{M}\left(1+\frac{1}{p_k}\right)\leq \zeta(2)\cdot C_M\tag{2} $$ is bounded by an absolute constant, depending on $M$ only.
But the LHS of $(2)$ is arbitrarily large, hence the set of prime numbers cannot be finite, qed.
Indeed $(2)$ can be used to prove a very weak form of the prime number theorem: $$ \sum_{\substack{p\text{ prime}\\p\leq N}}\frac{1}{p}\geq \log\log N-O(\log\log\log N).\tag{3}$$ About that, see Mertens' theorems.