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This is an extract from Paulo Ribenboim: 13 lectures on Fermats last theorem on page 105. "In 1845, von Staudt determined some factors of the numerator $N_{2k}$. Let $2k = k_1k_2$ with $gcd(k_1,k_2)= 1$ such that $p|k_2$ if and only if $p|D_{2k}$ then $k_1$|$N_{2k}$". Where $N_{2k}$ and $D_{2k}$ are the numerators and denominators of Bernoulli number $B_{2k}$.

So I've actually used the result of this theorem for some other proof, but looking back at it I find it is not true. For example when $2k=74$, then $2k=2\cdot37$. If we take $p=37$, we see that $37|k_2=37$ and so 37 must divide the denominator $D_{74}$ but $D_{74}=6$. I'm not sure what I'm missing here. Perhaps I have misinterpreted the theorem. Could someone clear this up for me?

hhhhh2hh
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  • The statement that is in quotation marks, it doesn't seem to hold, so my question is whether or not I've misinterpreted the statement. I have demonstrated what I believe the statement says via the example with 74. The reason why it is an issue is because Ribenboim actually uses this statement to prove that a prime is irregular if and only if there exists an integer k such that p divides $\frac{N_{2k}}{k_1}$. – hhhhh2hh Jul 31 '19 at 21:32
  • It can be found on the bottom of page 105 on http://staff.math.su.se/shapiro/ProblemSolving/13%20Lectures%20on%20Fermat%27s%20Last%20Theorem.pdf – hhhhh2hh Jul 31 '19 at 21:34
  • Please make it more clear and simple. What non-trivial property of $B_k$ would you like to understand, and what is its application. Also for the properties of $B_k$ there is Washington's book "cyclotomc fields" and the texts on $p$-adic L-functions or modular forms. – reuns Jul 31 '19 at 21:36
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    I do not think you have the complete statement should it not be " let $2k=k_{1}k_{2}$ with $gcd(k_{1},k_{2})=1$ such that $p|k_{2}$ if and only if $k_{2}|D_{2k} $then $k_{1}|N_{2k}$" in that case 2|6 and so 37 would divide the numerator –  Jul 31 '19 at 21:44
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    In ramanujans notebook part 1 entry 19(ii) the following statement appears :"The numerator of $B_{2n}$ is divisible by the largest factor of $2n$ which is relatively prime to the denominator of $B_{2n}$" Link to source :https://books.google.se/books?id=IJHaBwAAQBAJ&pg=PA123&lpg=PA123&dq=von+staudt+numerator&source=bl&ots=ndaHzb7xt1&sig=ACfU3U0Y_5jRwKeIzuT0rWzEdQAMIA71Lw&hl=sv&sa=X&ved=2ahUKEwiCiPDGkuDjAhXQJpoKHaUmDqoQ6AEwA3oECAMQAQ#v=onepage&q=von%20staudt%20numerator&f=false –  Jul 31 '19 at 22:03
  • You're right I didn't mention that part of the statement. But even with your example what is stopping me to choose $p=37$ then $k_2=37$. It is an if and only if statement so I imagine if I find a $p|k_2$ then $p$ must divide the denominator but it clearly depends on the choice of $k_1$ and $k_2$ which was not clear initially. Perhaps if I adjoin my statement with Ramanujans, then it is quite clear that we should choose $k_2$ to be 2. – hhhhh2hh Aug 01 '19 at 10:05
  • I believe that what the author meant was "any prime which divides $D_{2k}$ and $2k$ will dividide $k_{2}$ and $k_{2}$ is the factor of $2k$ for which any prime dividing $k_{2}$ will divide $D_{2k}$". But I do nevertheless agree with you that the wording is a bit confusing and I think the only thing that will completely settle it is to see the proof, which I have not been able to find. –  Aug 01 '19 at 18:27

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I can't explain what Ribenboim was trying to write, because it is not correct. Perhaps he meant to use the Von-Staudt Clausen theorem. The Wikipedia article states

Specifically, if $n$ is a positive integer and we add $1/p$ to the Bernoulli number $B_{2n}$ for every prime $p$ such that $p − 1$ divides $2n$, we obtain an integer [...]

This fact immediately allows us to characterize the denominators of the non-zero Bernoulli numbers $B_{2n}$ as the product of all primes $p$ such that $p − 1$ divides $2n$; consequently the denominators are square-free and divisible by $6$.

In your case of $2n=74$ the only divisors of $74$ are $2$ and $37$ and only the primes $p=2$ and $p=3$ are such that $p-1$ divides $74$. Hence $D_{74}=6.$

As user Armatowski commented, there is a result of Ramanujan that is close to what is stated by Ribenboim. This is from Bruce C. Berndt, Ramanujan's Notebooks, Part I, page 123.

Entry 19(ii). The numerator of $B_{2n}$ is divisble by the largest factor for $2n$ which is relatively prime to the denominator of $B_{2n}$.

$\quad$ Entry 19(ii) is contained in (18) of Ramanujan's paper [4] and is originally due to J. C. Adams. (See Uspensky and Heaslet's book [1, p.261].) In fact, in both Entry 19(ii) and (18) of [4], Ramanujan claims a stronger result, viz., the implied quotient is a prime number. However, this is false, for example, the numerator of $B_{22}$ is $854513=11\cdot131\cdot593.$

Somos
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  • I am familiar with this theorem, however it's not what I was looking for. But thanks anyways. – hhhhh2hh Jul 31 '19 at 20:31
  • @Somos, Can you answer the question-https://math.stackexchange.com/questions/3404245/how-to-associate-bernouli-numbers-or-bernouli-polynomials-into-the-relation-s – MAS Oct 23 '19 at 06:58