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I have read in some number theory books and in some online resources that it is known that there exist infinitely many irregular primes (a fact apparently proven quite some time ago, around 1915 by K. L. Jensen according to the Wikipedia entry).

I haven't been able to find any reference, either in books or in the internet as to what the method for proving this might have been, but in any case, what I was more curious is about the conjectured existence of infinitely many regular primes.

Even it is conjectured that there are "more" regular primes than irregular ones (about 61%), but the online references do not seem to say anything about its status, apart from saying that it is not known to be true.

Thus the questions I have are:

1) Are there any approaches at all to this problem?

2) Are there any other conjectures known to imply the existence of infinitely many regular primes?

3) Is it known why it is harder to prove this (other than the fact that it would give a proof of Fermat's Last Theorem for infinitely many prime exponents that does not involve heavy machinery =P) ?

Thank you very much in advance.

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    $\begingroup$ You can find a proof that there exist infinitely many irregular primes in Chapter 5 of Washington's book on cyclotomic fields. $\endgroup$ – Kevin Ventullo Feb 16 '11 at 9:40
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  1. and 2.: I am not aware of any ideas in this direction.

  2. The most obvious comment is the fact that there are Bernoulli numbers, and that their numerators must have prime factors. If these prime factors are large enough, they will produce irregular primes.

    For proving the existence of regular primes you have to show that there exist primes $p$ not dividing the numerators of the Bernoulli numbers $B_n$ with $n < p$. Heuristically there must be many of them, but the numerators grow so fast that you cannot exclude the possibility that all primes larger than some bound are irregular.

The only related result I am aware of is that the irregularity index cannot be too large. This was used in Ribet's famous proof of the converse of Herbrand's theorem, and was given a very simple proof by Metsänkylä.

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