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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 $\frac{2}{5}$: $\prod_p(\frac{p^2-1}{p^2+1})=\frac{3}{5} \cdot \frac{8}{10} \cdot \frac{24}{26} \cdot \frac{48}{50} \cdots$.

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    $\begingroup$ The result is attributed to Ramanujan (1913-1914), who may have used other methods (like hypergeometric series, continued fractions etc.). A reference is Le Lionnais, F. Les nombres remarquables. Paris: Hermann, p. 46, 1983. $\endgroup$ – Dietrich Burde Apr 23 '14 at 11:27
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    $\begingroup$ The product seems "simple" only because the expression was rigged to remove the powers of $\pi$ in $\zeta(2)$ and $\zeta(4)$. Also, your factorization with terms $(p-1)/(p\pm i)$ is into two products that are each divergent, though that wouldn't stop Ramanujan. $\endgroup$ – KConrad Apr 23 '14 at 11:48
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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.

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    $\begingroup$ @ David Speyer: No, I haven't seen it. It seems slightly related to the comment by i707107 on my answer to the MO question"What are the connections between pi and prime to numbers " just over a year ago, though that comment does not evaluate the product independently as far as I can see. It would be ineteresting to know if your method generalizes to an elementary proof that $\zeta(2m)^{n}/\zeta(2n)^{m}$ is always rational. $\endgroup$ – Geoff Robinson May 30 '14 at 19:39
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    $\begingroup$ Re. your write-up (which I have only scanned very quickly): Your "dropping the mask" section (especially the parenthetical remark) is excellent. It would be good if more papers included such an explanation of where reasoning has come from. $\endgroup$ – Nick Gill May 30 '14 at 20:38
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    $\begingroup$ I think you should totally send your write-up to the Monthly! $\endgroup$ – Greg Martin May 30 '14 at 23:39
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    $\begingroup$ @DavidSpeyer : I edited your punctuation a bit and I think you should do the same in the pdf document that you link to. $\endgroup$ – Michael Hardy Jun 1 '14 at 19:51
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    $\begingroup$ @VladimirDotsenko Yes it is. That's why I didn't wind up sending this anywhere. See testcase's answer below. $\endgroup$ – David E Speyer Mar 27 '15 at 14:02
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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.

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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.

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    $\begingroup$ The actual reference being "Quelques conséquences surprenantes de la cohomologie de SL2(Z)" in Leçons de Mathématiques d'aujourd'hui, Vol. 1,Cassini, Paris (2000). 99-123. An earlier paper by Zagier also mentionning this is "Values of zeta functions and their applications", in First European Congress of Mathematics, Volume II, Progress in Math. 120, Birkhäuser-Verlag, Basel, (1994) 497-512, people.mpim-bonn.mpg.de/zagier/files/scanned/… $\endgroup$ – Thomas Sauvaget Jun 1 '14 at 13:04
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    $\begingroup$ Thanks, this is really interesting. It looks like Zagier returned to this again in "PERIODS OF MODULAR FORMS, TRACES OF HECKE OPERATORS, AND MULTIPLE ZETA VALUES" people.mpim-bonn.mpg.de/zagier/files/kokyuroku/843/fulltext.pdf and he cites an article of Skoruppa sciencedirect.com/science/article/pii/S0022314X83710073 which proves identities between Eisenstein series in a similar manner. On a quick glance, it looks to me like Zagier is using Taylor series of elliptic functions where I am using Taylor series of trig functions. But I need to read more closely. $\endgroup$ – David E Speyer Jun 2 '14 at 2:51
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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).

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