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How to prove that $p-1$ is squarefree infinitely often, where $p$ is prime?

I had thought of using Dirichlet's theorem on arithmetic progressions. The number of squareful values of $p-1$ less than or equal to $n$ is bounded by $\sum_{q}{\pi_{q^2,1}(n)}$, each term $\pi_{q^2,1}(n) \approx \frac{\pi(n)}{q \cdot (q-1)}$, and $\sum_{q}{\frac{1}{q \cdot (q-1)}} \lt 1$. But using for example this statement of Dirichlet's theorem, the possibility is open that the error in each approximation for $\pi_{q^2,1}(n)$ may continue to increase with $q$, and since the sum runs up to $\sqrt{n}$, this argument is invalid.

Is there a more refined version of Dirichlet's theorem that can be used to circumvent this issue? Or an entirely different way to prove this statement?

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If you're willing to accept a certainly-true-but-unproved hypothesis, there are infinitely many primes $q$ such that $p=2q+1$ is prime, and those $p$ will work. –  Gerry Myerson Jun 3 '12 at 4:53
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I'm hoping for an unconditional proof (or else a reference to the problem being open). Infinity of safe primes is open and implies it but I don't know why this should be as hard. Are you saying RH implies an infinity of safe primes? I hadn't heard of this and I looked for a reference on Google but couldn't find one. However, I can see that RH implies that $p-1$ is squarefree infinitely often if the error is independent of $q$. I'm not very familiar with the proof of Dirichlet's theorem (I looked at it but found it quite intimidating) so I don't know if this is true. –  Dan Brumleve Jun 7 '12 at 3:24
    
The "certainly-true-but-unproved hypothesis" I had in mind was not that of Riemann, but of Dickson; see, e.g., en.wikipedia.org/wiki/Dickson's_conjecture. –  Gerry Myerson Jun 7 '12 at 5:51
    
For what it's worth, these primes are tabulated at oeis.org/A039787, but no other information is given there. –  Gerry Myerson Jun 7 '12 at 5:58

2 Answers 2

up vote 6 down vote accepted
+100

There is a more refined version of Dirichlet's theorem which is immensely useful in these types of problems. It's called the Bombieri-Vinogradov theorem, which says roughly that the error term in Dirichlet's theorem cannot be very large for many values of $q$ simultaneously. While we can't bound any individual error term by $O(n^{1/2+\epsilon})$ (such a bound would be equivalent to a form of RH), we can get this level of control if we average over many values of $q$.

More precisely, let $E(q)$ denote the error $\left|\pi_{q^2,1}(n) - \frac{\pi(n)}{q(q-1)}\right|$ in Dirichlet's theorem. Then a simple consequence of Bombieri-Vinogradov is that for some constant $B > 0$, $$\sum_{q < n^{1/4} \ \log^{-B} n} E(q) \ll \frac{n}{\log^2n}.$$

Therefore the error terms for $q \le n^{1/4}\ \log^{-B} n$ cannot accumulate to exceed the $\Omega(n/\log n)$ difference between $\pi(n)$ and $\sum\limits_{q\text{ prime}} \frac{\pi(n)}{q(q-1)}$, exactly as you had hoped. The only moduli remaining are $n^{1/4} \ \log^{-B} n \le q \le \sqrt{n}$, but these are very easily controlled using the trivial bound $\pi_{q^2,1}(n) \le \frac{n}{q^2}$ (there aren't many numbers congruent to $1\!\!\pmod{q^2}$ when $q$ is large).

A slightly more general application of Bombieri-Vinogradov should be able to produce the asymptotic result cited in @GerryMyerson's answer.

EDIT — Had the wrong upper bound for $q$, since the modulus is $q^2$, not $q$.

EDIT 2012/06/13: The asymptotic is probably not as immediate as I thought; it would require some subtlety with regard to the number of primes $q$ we can sieve out.

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It appears that a density result for these primes was given by Mirsky, with a proof appearing in a paper by Moree and Hommerson. Section 6.2.1, Theorem 2 (page 17) says if $r$ is a non-zero integer, $k$ an integer greater than 1, $H$ any positive number, then the number of primes $q$ up to $x$ such that $q-r$ is $k$-free is $$\prod_{p\nmid r}\left(1-{1\over p^{k-1}(p-1)}\right){\rm Li}(x)+O\left({x\over\log^Hx}\right)$$

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