Let $n>1$ be an integer, and let us consider the set $P(n)$ of all prime numbers $p$ such that $p$ is not congruent to $1$ modulo $n$. Dirichlet's Density Theorem tells us that $P(n)$ has a natural density, equal to $$1-\varphi(n)^{-1}$$ where $\varphi(n) = |(\mathbb Z /n)^\ast|$ is Euler's totient.

From the Frobenian point of view, saying that $p$ is congruent to $1$ modulo $n$ is to say that the ideal $(p)$ splits completely in the cyclotomic field $\mathbb Q(\zeta_n)$.

From Chebotarev's point of view, saying that $p$ is congruent to $1$ modulo $n$ is to say that the Frobenius element over $p$ in $\operatorname{Gal}(\mathbb Q(\zeta_n)|\mathbb Q) \simeq (\mathbb Z /n)^\ast$ is the identity.

So far so good, now let us consider the set $P$ of all prime numbers $p$ which are not congruent to $1$ modulo $n^2$ for any $n>1$, that is $$P := \bigcap_{n>1}P(n^2) = \bigcap_{\ell\mathrm{ prime}}P(\ell^2)$$ Supposing that "the events $P(\ell^2)$ are uncorrelated" for different $\ell$'s, we can phantasise about the density of $P$, hoping it might be (at least up to a rational factor, I don't vouch for it) $$\operatorname{dens}(P) = \prod_\ell 1-\frac{1}{\ell(\ell-1)} \quad = 0.37395581361920228805...$$ a number called Artin's constant (it appears in Artins primitive root conjecture, which is similar in nature). The question whether $P$, or similarly constructed sets of primes, have a density and whether it is the expected one goes far beyond the density theorems of Dirichlet, Frobenius and Chebotarev. The corresponding Galois extension would be the maximal cyclotomic extension of $\mathbb Q$, which is ramified everywhere.

Can you name this problem? Have you seen it before? Where?

Hooley (1967) has shown that Artins primitive root conjecture follows from GRH. In principle, the problem of determining the density of $P$ should be simpler.

Under GRH, is it true that the density of $P$ exists and is equal to Artin's constant?

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    $\begingroup$ Interesting question, but perhaps a more precise title is appropriate. $\endgroup$ Aug 5, 2011 at 16:03

2 Answers 2


The specific density result you quote is a result of Mirsky, see

L. Mirsky, "The number of representations of an integer as the sume of a prime and a k-free integer", Amer. Math. Monthly 56 (1949)

There have been several generalizations, for the direction on replacing squares with higher powers see "Values of the Euler function free of k-th powers" by W.D. Banks and F. Pappalardi. For the result on primes not congruent to $\frac{a}{b}\pmod{n^2}$ see "Arithmetic progressions, prime numbers, and squarefree integers" by S. Clary and J. Fabrykowski.


Let me sketch a proof. If you fix a bound $z$, then the events $P(\ell^2)$ for different $\ell \leq z$ are uncorrelated; this is just a consequence of the prime number theorem for progressions and the multiplicativity of Euler's function. This reduces the problem to "understanding the tails"; in other words, we have to show that as $z\to\infty$, the relative upper density of the primes divisible by $\ell^2$ for some $\ell > z$ tends to zero.

Consider the primes $p \leq x$. By the Brun--Titchmarsh inequality, the number of such $p$ for which $p-1$ is divisible by $\ell^2$ for some prime $\ell$ with $z < \ell \leq x^{1/4}$ (say) is $$ \ll \sum_{z< \ell \leq x^{1/4}} \frac{x}{\phi(\ell^2)\log{(x/\ell^2)}} \ll \pi(x) \sum_{z > \ell} \frac{1}{\ell^2} \ll \frac{\pi(z)}{x}. $$ Also, the number of $p$'s with $p-1$ divisible by $\ell^2$ for some $\ell > x^{1/4}$ can be estimated trivially: We just count how many $n \leq x$ are divisible by some $\ell^2$ with $\ell > x^{1/4}$, which is clearly at most $\sum_{\ell > x^{1/4}} \lfloor x/\ell^2\rfloor \ll x^{3/4}$. This is negligible for us. Hence, the relative upper density in the previous paragraph is $\ll 1/z$. So it does indeed tend to zero as $z\to\infty$.


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