Infinite Regular Primes Conjecture

Another unproven conjecture in number theory is the Infinite Regular Primes Conjecture (IRPC). It addresses a special properties of primes. A prime can be regular or a prime can be irregular. For the later we know a proof [given below] since many years. For regular primes it is still unknown if they are infinite despite the fact that we can give good arguments that around $$60.653\%$$ of all primes should be regular. So we know that major part of all primes should be regular, but we cant prove it.
A regular prime can be characterised in more than one way.

Definition [Regular Prime]

A prime \(p > 2\) is called regular if it does not divide the class number of the \(p\)-th cyclotomic field $\blacktriangleleft$.

An alternative characterisation is the following.

Definition [Regular Prime]

A prime \(p > 2\) is called regular if it does not divide the numerator of any Bernoulli number \(B_n\) for \(n = 2,4,6,...,p-3\) $\blacktriangleleft$.

Then the IRPC conjecture simply this:

Infinite Regular Primes Conjecture

There are infinite many regular primes.

Kummer proved in 1850 that Fermat's Last Theorem is true if the involved exponent $p$ is regular. The infiniteness of irregular primes has already been proved several decades ago (1954) by Carlitz [1]. His proof is based on the second characterisation from above. It is a proof by contradiction. He assumes that there are only finite many irregular primes, which leads him to a contradiction to some proven result about Bernoulli numbers.

Proof (Infinite Irregular Primes). [Following Carlitz [1]]. For the proof Carlitz uses several properties of the Bernoulli numbers:

1. \(B_m \equiv 0\pmod{p^r}\) if \(p^r | m\) and \((p-1) \nmid m\)
2. \(pB_m \equiv -1\pmod{p}\) if \((p-1)|m\)   (Staudt-Clausen Theorem)
3. \(\frac{B_{m+r(p-1)}}{m+r(p-1)} \equiv \frac{B_m}{m}\pmod{p}\)   (Kummer's Congruence)

Contradiction assumption: The number of irregular primes is finite, say \(p_1,p_2,...,p_k\).

We build the integer \(M = 2^t\prod^k_{i=1}(p_i-1)\), for an arbitrary integer \(t\) that we need later. The denominator of the \(m\)-th Bernoulli number is defined as \(\prod_{(p-1)|m}p\). We build the quotient $$ \frac{B_M}{M} = \frac{N_M}{M\cdot D_M} $$ whereof \(N_M\) is the numerator of the \(M\)-th Bernoulli number and \(D_M\) its denominator. If we plug in in the values for \(M\) and \(D_M\) we get $$ \frac{B_M}{M} = \frac{N_M}{2^t\prod^k_{i=1}(p_i-1)\cdot \prod_{(p-1)|M}p}$$ Note that in the denominator \(D_M = \prod_{(p-1)|M}p\) all irregular primes are contained. Now take item 2 from above, i.e., $$p_iB_M = \frac{N_M}{\prod_{(p-1)|M}p} \equiv -1\pmod{p_i}$$ which holds for every irregular prime \(p_i\). It follows, since non of them yields zero, that \(N_M\) does not contain any of the primes \(p_i\). Thus the only primes that can be contained in \(N_M\) are, according to item 1., factors of \(M\). So the new numerator \(N'_M\), after cancellation with \(M\) is

$$ \frac{B_M}{M} = \frac{N_M}{M\cdot D_M} = \frac{N'_M}{D'_M}$$

and $N'_M$ must be equal to $\pm 1$, since no irregular prime is contained and all other possible factors have been cancelled out by division with \(M\) $$ \frac{B_M}{M} = \pm \frac{1}{D'_M}$$ The contradiction now occurs, because it is well known that

$$ \lim_{m \rightarrow \infty}{\frac{B_m}{m}} = \infty$$

However, the \(t\) in the definition of \(M\) could be arbitrary large. So one could use this freedom to establish.

$$ \lim_{M \rightarrow \infty}{\frac{B_M}{M}} = \lim_{M \rightarrow \infty}{\frac{1}{D'_M}} = 0$$

, so the assumption that there are only finite irregular primes is wrong.

Q.e.d.

If one wants to use a similar approach to prove that there are infinite regular primes, he will probably fail, since he could not construct an object (like the \(M\)-th Bernoulli number) that contains all finite regular primes in a similar easy way.

Heuristics arguments also suggest that the fraction of regular prime is even larger. It is estimated that around \(60.64\%\) are regular.

Based on calculation from the blog post The Agoh-Guiga Conjecture, one could give another characterisation of a regular prime

Definition [Regular Prime, Reformulation]

An odd prime \(p\) is regular if there is no even number \(1 \leq n \leq p-3\) such that $$\sum^{(p-1)/2}_{k=1}k^{n-1} \equiv 0\pmod{p}\;\text{and}\; p\nmid (2^n-1)$$

Proof. Since it holds that (see The Agoh-Guiga Conjecture) $$\sum^{(p-1)/2}_{k=1}k^{n-1} \equiv n^{-1}2^{-n}2(1-2^n)B_n \pmod{p}$$ the term can only be zero if either $(1-2^n)$ is divisible by $p$ or $B_n$. Since we excluded the former per definition the claim follows.
Q.e.d.

Note that a regular prime number $p$ with $p | (1-2^n)$ (as well as $B_n$) is not covered by the reformulation.
If we neglect the requirement that $p\nmid (2^n-1)$ for the moment. Is it easier to prove that there exists infinite many primes with the property that $$\sum^{(p-1)/2}_{k=1}k^{n-1} \not\equiv 0\pmod{p}$$ for all even $n \leq p-3$?
That means, one could perhaps show that for infinitely many primes
$$ \prod^{(p-3)/2}_{j=1}\sum^{(p-1)/2}_{k=1}k^{2j+1} \not\equiv 0\pmod{p}$$ hold.

Also the density heuristic can be easily derived from the reformulation. The sum $\sum^{(p-1)/2}_{k=1}k^{n-1}$ should be not equal to zero, for all even $n$ less than $(p-3)/2$. Hence $(p-1)/p$ values are valid sum results. If we consider this for all $(p-3)/2$ even integer less than $p-3$ , we have
$$ \lim_{p \rightarrow \infty}{\prod^{(p-1)/3}_{j=1}\frac{p-1}{p}} = \left(1 - \frac{1}{p}\right)^{(p-3)/2} \approx \frac{1}{\sqrt{e}} \approx 0.60653$$
which yields also around $60.65\%$.

[1] Carlitz, L. (1954). "Note on irregular primes". Proceedings of the American Mathematical Society (AMS) 5: 329–331