# Can someone explain this theorem from von Staudt on denominators of Bernoulli numbers

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?

• 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 staff.math.su.se/shapiro/ProblemSolving/… – 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
• 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 – André Armatowski Jul 31 '19 at 21:44
• 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 :books.google.se/… – André Armatowski Jul 31 '19 at 22:03

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 $$854513=11\cdot131\cdot593.$$

• 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-math.stackexchange.com/questions/3404245/… – Why Oct 23 '19 at 6:58