Are there infinitely many primes that are 1 more a square-free number?

Question

Other words, are there infinitely many primes $p$ such that $p-1$ is square-free. This seems likely to be true, but I can't seem to give an easy argument why. This set of primes and all primes 1 more an integer with a square divisor make up the set of primes, so at least one of these is infinite. I looked at this question for enlightenment but to no avail : Infinitely many primes $p$ such that $\frac{p-1}{2}$ is a product of two primes as 2 could be included in this product.

Answer 1

It is in fact very simple to count all such primes. One has study the sum $$\sum_{p\leq x} \mu(p-1)^2.$$ By inversion it equals $$\sum_{1\leq d \leq \sqrt{x}} \mu(d) \#\{p\leq x : p \equiv 1\mod{d^2 } \} .$$ The cardinality is at most $x/d^2$, hence the contribution of $d>(\log x)^{100} $ is $$\ll x\sum_{d>(\log x)^{100} } 1/d^2 \ll x (\log x)^{-100}.$$ By Siegel--Walfisz we see that $$\sum_{1\leq d \leq (\log x)^{100}} \mu(d) \#\{p\leq x : p \equiv 1\mod{d^2 } \} =\frac{x}{\log x} \sum_{1\leq d \leq (\log x)^{100}} \mu(d)\phi(d^2)^{-1} +O\left(\frac{x}{(\log x )^{100}}\right) .$$ The standard bound $\phi(m)\gg m^{1/2}$ now yields $$\sum_{1\leq d \leq (\log x)^{100}} \mu(d)\phi(d^2)^{-1}=\sum_{d\in \mathbb{N}} \mu(d)\phi(d^2)^{-1}+O((\log x)^{-3}).$$ Putting everything together shows that $$\sum_{p\leq x} \mu(p-1)^2=c \frac{x}{\log x} +O\left(\frac{x}{(\log x)^2 }\right),$$ where $$ c=\sum_{d\in \mathbb{N}} \mu(d)\phi(d^2)^{-1}=\prod_p \left(1+\frac{1}{p(p-1)}\right).$$ The last expression shows that $c>0$ , hence there are infinitely many primes $p$ such that $p-1$ is square-free.