Computing $\prod_p(\frac{p^2-1}{p^2+1})$ without the zeta function?

Question

We see that $$\frac{2}{5}=\frac{36}{90}=\frac{6^2}{90}=\frac{\zeta(4)}{\zeta(2)^2}=\prod_p\frac{(1-\frac{1}{p^2})^2}{(1-\frac{1}{p^4})}=\prod_p \left(\frac{(p^2-1)^2}{(p^2+1)(p^2-1)}\right)=\prod_p\left(\frac{p^2-1}{p^2+1}\right)$$

$$\implies \prod_p \left(\frac{p^2-1}{p^2+1}\right)=\frac{2}{5},$$

But is this the only way to compute this infinite product over primes? It seems like such a simple product, one that could be calculated without the zeta function.

Note that $\prod_p(\frac{p^2-1}{p^2+1})$ also admits the factorization $\prod_p(\frac{p-1}{p-i})\prod_p(\frac{p+1}{p+i})$. Also notice that numerically it is quite obvious that the product is convergent to $\frac{2}{5}$: $\prod_p(\frac{p^2-1}{p^2+1})=\frac{3}{5} \cdot \frac{8}{10} \cdot \frac{24}{26} \cdot \frac{48}{50} \cdots$.

Answer 1

I found a proof of $$5 \sum_{m=1}^{\infty} \frac{1}{m^4} = 2 \left( \sum_{n=1}^{\infty} \frac{1}{n^2} \right)^2$$ by rearranging sums and wrote it up. The argument is just 1.5 pages, the other 4.5 are explanations and context.

Here is a summary using divergent sums; see the write up for a correct version. Set $$h(m,n) = \begin{cases} \frac{1}{m^3 (n-m)} & m \neq n,\ m \neq 0, \\ 0 & m=n \ \text{or} \ m=0. \end{cases}$$ Then we should have $$\sum_{(m,n) \in \mathbb{Z}^2} h(m,n) - h(n,2n-m) =0$$ as every value occurs twice with opposite signs. So $\sum_{(m,n) \in \mathbb{Z}^2} g(m,n)=0$ where $$g(m,n) := h(m,n) - h(n,2n-m) = \begin{cases} \frac{m^2+mn+n^2}{m^3 n^3} & m \neq n,\ m, n \neq 0 \\ - \frac{1}{m^4} & n=0,\ m \neq 0 \\ - \frac{1}{n^4} & m=0,\ n \neq 0 \\ 0 & m=n \end{cases}$$ Group together the terms where $(|m|, |n|)$ have a common value; we get $\sum_{(m,n) \in \mathbb{Z}_{\geq 0 }^2} f(m,n) =0$ where $$f(m,n) = \begin{cases} \frac{4}{m^2 n^2} & m \neq n,\ m,n >0, \\ - \frac{2}{m^4} & m>n=0, \\ - \frac{2}{n^4} & n>m=0, \\ - \frac{2}{m^4} & m=n>0, \\ 0 & m=n=0. \end{cases} $$ Writing this out, $4\zeta(2)^2 - 6 \zeta(4) - 4 \zeta(4)=0$, as desired.

Has anyone seen this? If this is new, I'm thinking of sending it to the Monthly.

Answer 2

A proof of this identity not using properties of the Riemann zeta function is listed as an unsolved problem in section B48 of Guy's Unsolved Problems in Number Theory.

An amusing observation: this identity implies the infinitude of primes. If the product over $p$ were finite, the final answer would be a rational number with a factor of $3$ in its numerator --- the very first term in the product has a $3$ in the numerator, and $3$ cannot divide $u^2+1$ for any integer $u$, so the $3$ can never cancel.

Answer 3

An elementary proof of the identity $$ 2 \zeta(2)^2 = 5\zeta(4)$$ has been found by Don Zagier in the paper

http://people.mpim-bonn.mpg.de/zagier/files/tex/ConsequencesCohomologySL/fulltext.pdf

The idea seems similar to how David Speyer proves it in that he starts by defining the function

$$ f(m,n) = \frac{2}{n^3m} + \frac{1}{n^2m^2} + \frac{2}{nm^3} $$ and then verifies that $$ f(m,n) -f(m,n+m) - f(m+n,n) = \frac{2}{m^2n^2}.$$

But since we also have $$ \sum_{n,m>0} f(m,n) - \sum_{n,m>0} f(m,n+m) - \sum_{m,n>0} f(m+n,n) = \sum_{n>0} f(n,n),$$ since only the diagonal terms survive, this gives the identity $$ 2 \zeta(2)^2 = 5\zeta(4).$$

The crucial function $f(m,n)$ seems to be part of a larger family of functions with ties to period polynomials, as explained in the paper.

Answer 4

I would guess that every introductory paper about polynomial relations of multiple zeta values (from Kontsevich-Manin to F. Brown) explains this identity in an elementary manner. But I have no precise reference to offer now, except for E. Panzer's notes (see the formula on top of page 4).