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.

  • $\begingroup$ I suspect there's likely no known way to prove this, just like with so relatively many other fairly simple to state conjectures involving primes. In case you haven't thought about this, one possibility would be if anybody could prove there's an infinite number of primes of the form of a primorial plus $1$, i.e., $p_n\# + 1$. The first few such primes, & other details, are listed in the OEIS A014545. $\endgroup$ – John Omielan Jun 8 at 0:25
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    $\begingroup$ Another area of relevance are the Sophie Germain primes, which have the property that $2p+1$ is also a prime. Obviously, $2p$ is square-free. The derived prime $2p+1$ is called a safe prime. It is hypothesized, but not proven, that there are infinitely many Sophie Germain primes. $\endgroup$ – Keith Backman Jun 8 at 1:05
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    $\begingroup$ I would suspect that there are infinitely many $n$ such that $p_1p_2\dots p_n+1$ is prime. I don't think there is a proof though. $\endgroup$ – Julian Mejia Jun 8 at 4:03
  • $\begingroup$ @JulianMejia FYI, a Primorial is $p_n\# = p_1p_2\ldots p_n$, as I discussed in my earlier comment. $\endgroup$ – John Omielan Jun 8 at 4:20
  • $\begingroup$ Exception is for Format primes $P_F=2^{2^n}+1$ because $2^{2^n}$ is square.Also Euler's old question is : are number of primes of the form $x^2+1$ limited? which has not been answered yet. I dont think there are more exceptions. $\endgroup$ – sirous Jun 8 at 15:09

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.

  • $\begingroup$ Nice! Never heard of Siegel--Walfisz... Guess I gotta study more. Thanks! $\endgroup$ – Sean Nemetz Jun 11 at 2:51
  • $\begingroup$ For the inversion if I take $x=6$, I get $-1$. So this can't be equal to our original sum? $\endgroup$ – Sean Nemetz Jun 11 at 5:00

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