# Irregular prime number

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An odd prime number $p$ for which the number of classes of ideals in the cyclotomic field $\mathbf Q(e^{2\pi i/p})$ is divisible by $p$. All other odd prime numbers are called regular.

Kummer's test allows one to solve for each given prime number the problem of whether it is regular or not: For an odd prime number to be regular it is necessary and sufficient that none of the numerators of the first $(p-3)/2$ Bernoulli numbers $B_2,B_4,\dots,B_{p-3}$ is divisible by $p$ (cf. [1]).

The problem of the distribution of regular and irregular prime numbers arose in this connection. Tables of the Bernoulli numbers and Kummer's test indicated that among the first hundred there are only three irregular prime numbers, 37, 59, 67 (the numerators of $B_{32}$, $B_{44}$ and $B_{58}$ are multiples of 37, 59 and 67, respectively). E. Kummer conjectured that there are on the average twice as many regular prime numbers as irregular ones. Later C.L. Siegel [2] conjectured that the ratio of irregular prime numbers to regular prime numbers contained in an interval $(1,x)$ tends to $e^{-1/2}$ as $x\to\infty$ (here $e$ is the base of natural logarithms). Up till now (1989) it is only known that the number of irregular prime numbers is infinite. There are 439 regular and 285 irregular prime numbers among the odd numbers smaller than 5500, cf. [3].

For any regular $p$ the Fermat equation

$$x^p+y^p=z^p$$

does not have non-zero solutions in the rational numbers [1].

Let $p$ be an irregular prime number, let $2\alpha_1,\dots,2\alpha_s$ be the indices of the Bernoulli numbers among $B_2,B_4,\dots,B_{p-3}$ whose numerators are divisible by $p$ and let $k$ and $t$ be natural numbers such that $q=1+pk$ is a prime number smaller than $p(p-1)$ and $t^k\not\equiv1\bmod q$. Let

$$d_\alpha=\sum_{r=1}^{(p-1)/2}r^{p-2\alpha},$$

$$D_\alpha=t^{-kd/2}\prod_{r=1}^{(p-1)/2}(t^{kr}-1)^{r^{p-1-2\alpha}}.$$

If for each $\alpha=\alpha_i$, $i=1,\dots,s$,

$$D_\alpha^k\not\equiv1\pmod q,$$

then for the irregular prime number $p$ Fermat's theorem holds, i.e. the Fermat equation is unsolvable in the non-zero rational numbers. This is called Vandiver's test. The truth of Fermat's theorem for all exponents smaller than 5500 has been proved by using Vandiver's test (cf. [4]).

#### References

 [1] E. Kummer, "Allgemeiner Beweis des Fermatschen Satzes, dass die Gleichung $x^\lambda+y^\lambda=z^\lambda$ durch ganze Zahlen unlösbar ist, für alle diejenigen Potenz-Exponenten $\lambda$, welche ungerade Primzahlen sind und in den Zählern der ersten $\frac12(\lambda-3)$ Bernoullischen Zahlen als Factoren nicht vorkommen" J. Reine Angew. Math. , 40 (1850) pp. 130–138 [2] C.L. Siegel, "Zu zwei Bemerkungen Kummers" Nachr. Akad. Wiss. Göttingen Math. Phys. Kl. , 6 (1964) pp. 51–57 [3] Z.I. Borevich, I.R. Shafarevich, "Number theory" , Acad. Press (1966) (Translated from Russian) (German translation: Birkhäuser, 1966) [4] H.S. Vandiver, "Examination of methods of attack on the second case of Fermat's last theorem" Proc. Nat. Acad. Sci. USA , 40 : 8 (1954) pp. 732–735

The truth of Fermat's theorem has been established for all exponents $p<125000$ by S. Wagstaff [a1].
His computations show that $60.75\%$ of the 11733 odd prime numbers smaller than $125000$ are regular. This is in close agreement with Siegel's conjecture, which expects $e^{-1/2}\cong60.65\%$ of all prime numbers to be regular.
More generally, one defines the index of irregularity $i(p)$ of an odd prime number $p$ as the number of indices $k\in\lbrace2,4,\dots,p-3\rbrace$ for which $p$ divides the numerator of the Bernoulli number $B_k$. The regular prime numbers are the prime numbers $p$ satisfying $i(p)=0$. Heuristically, one expects the fraction of prime numbers $p$ for which $i(p)=k$ to be $(1/2)^ke^{-1/2}/k!$, and this is confirmed by the data in [a1]. It was proved by Eichler that the first case of Fermat's theorem holds for a prime exponent $p$ when $i(p)<\sqrt p-2$ (cf. [a2]). See also Fermat great theorem.