Factoring Zeta 

We've seen how Leibniz might have anticipated Euler by discovering that the sum of the inverse squares is π^{2}/6 purely on the basis of the arctangent series for π/4. Another way of expressing that approach is to regard it as essentially a factorization of the product form of ζ(2). In other words, beginning with the expression 



we factor the right hand side into two parts 



However, the two individual products on the right side aren't very useful because the first is just the harmonic series, which diverges to infinity, and the second converges on zero. In order to get a more useful factorization, we note that the above is really just a special case of a more general form 



where we've replaced the unit numerators with the functions a(p) and b(p), which for each prime p are the roots of x^{2} − 1, but any permutation of those roots is allowable for any given prime. For example, we are free to define a(p) and b(p) in terms of the Legendre symbol 



in which case we have decomposed ζ(2) into the factors 



where the first factor is simply 1/(1  1/2) times the arctan series for pi/4, and the second factor is 1/(1 + 1/2) times the signreversal of the Leibniz series (product) as discussed in the previous note. 

Now, what happens if we take this same approach to ζ(3), i.e., the sum of the inverse cubes? There is no known "simple" expression for ζ with odd arguments, in contrast with the fact that all the evenindex ζ values are given by 



where B_{2n} is the (2n)th Bernoulli number. The problem with this formula relative to odd ζ arguments is that all the oddindex Bernoulli numbers vanish. (Of course, the oddindex Bernoulli polynomials do not vanish, so people have tried to generalize the above expression in terms of the full polynomials rather than just the constant coefficients, but nothing seems to work.) 

Following our factorization of ζ(2), we might try the same trick on the productform of ζ(3), i.e., we might try to factor the product 



which can obviously be formally factored as 



but again we have the problem that the first factor is the harmonic series (divergent) and the second factor converges to zero. We might simply replace the "1/p" terms with a(p)/p and b(p)/p, but it seems more in the spirit of our previous example to factor ζ(3) into linear factors. 

To simplify the notation, let me use square brackets to denote a product evaluated over all primes, so our putative factorization of ζ(3) can be written as 



where a(p), b(p), and c(p) are now (some permutation of) the roots of x^{3} − 1. Except for the prime 3, every prime falls in one of the congruence classes {1,2}, {4,5}, {7,8} modulo 9, so if we let r_{1}, r_{2}, r_{3} denote the cube roots of 1 



then one possible way of defining the functions a,b,c is presented below, along with the resulting factorization of ζ(3). 



Multiplying the factors together gives 



Unfortunately, unlike the case of ζ(2), these "linear factors" aren't immediately recognizable as fractions of π (or anything else). However, there were quit a few arbitrary choices made in the above construction, mainly involving the definitions of the functions a,b,c. In the case of ζ(2) we essentially classified the primes into two categories according to whether x^{2} + 1 splits or is irreducible in the field of integers modulo p (i.e., whether or not −1 is a quadratic residue), so it would seem that for ζ(3) we should classify the primes into three categories according to whether some cubic polynomial splits completely, has one linear factor, or is irreducible in Z_{p}. However, we can't use x^{3} + 1 because that always has at least one linear factor. It's not clear to me how to come up with the "most natural" factorization of ζ(3). 

By the way, it's possible that we should really work with 3! = 6 factors, maybe using ± the cube roots of 1. This may be related to the fact that we can get ζ(3) from either of the products 1/(1 − 1/p^{3}) or 1/(1 + 1/p^{3}), because they are "conjugate" factors of ζ(6), i.e., 



So if we had a simple expression for either of the factors we would automatically have one for the other factor. Unfortunately the second factor is just as intractable as the first, so this doesn't seem to help much, except perhaps to suggest that the cube roots of −1 should also be included in a sixterm factorization. 
