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In order to know more about product over primes ,I would like to know how do I show that :$$\prod\frac{p^2+1}{p^2-1}=\frac{5}{2}$$ without using properties of Riemann zeta function ?

Note01 : it is well known that $$\prod\frac{p^2+1}{p^2-1}=\frac{{\zeta}^2(2)}{\zeta(4)}=\frac{5}{2}$$ but is there other method to show that ?

Note 02 :I wish using divisor function properties

Thank you for any help

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    $\begingroup$ See mathoverflow.net/questions/164092 $\endgroup$
    – David E Speyer
    Commented Jan 19, 2016 at 23:44
  • $\begingroup$ @nice proof ,thank you for your help $\endgroup$
    – zeraoulia rafik
    Commented Jan 19, 2016 at 23:46
  • $\begingroup$ @PaceNielsen,in mathoverflow.net/questions/164092 used Riemann zeta function properties , but i seek to use other properties as example :power sum divisor function ,all method which are different from zeta function properties uses !!! $\endgroup$
    – zeraoulia rafik
    Commented Jan 20, 2016 at 1:47
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    $\begingroup$ @zeraouliarafik I don't see any use of properties of the zeta function, unless you count using the fundamental theorem of arithmetic to get the Euler product. Are you seeing something else? $\endgroup$
    – Pace Nielsen
    Commented Jan 20, 2016 at 2:39
  • $\begingroup$ This is probably nonsense but: If you formally expand the factors of this product as power series in $p$, use the fundamental theorem of arithmetic, and make the heroic assumption that the product of infinitely many $-1$'s is $+1$, then the expression becomes $\sum_{n=1}^\infty 2^{\Omega(n)}n^2 = 5/2$, where $\Omega(n)$ is the number of distinct prime factors in $n$. I wonder if this could be true in some sense. For example, set $Z(s)=\sum_{n=1}^\infty 2^{\Omega(n)}n^{-s}$ where this converges, and analytically continue. Then we could hope that $Z(-2)=5/2$. Could this be? $\endgroup$
    – Steven Landsburg
    Commented Jan 21, 2016 at 4:42

2 Answers 2

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This is a well-known problem, attributed to Sam Wagstaff in Richard Guy's Unsolved Problems in Number Theory. Section B48 "Products taken over primes" includes a paragraph

Wagstaff asked for an elementary proof (e.g., without using properties of the Riemann zeta-function that $$\prod_p \frac{p^2+1}{p^2-1} = \frac52 $$ where the product is taken over all primes. It seems very unlikely that there is a proof which doesn't involve analytical methods. At first glance it might appear that the fractions might cancel, but none of the numerators are divisible by 3 [. . .]

with a reference to

David Borwein & Jonathan M. Borwein, On an intriguing integral and some series related to $\zeta(4)$, Proc. Amer. Math. Soc., 123 (1995) 1191-1198; MR 95e:11137.

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  • $\begingroup$ oh, sorry for this $\endgroup$
    – zeraoulia rafik
    Commented Jan 19, 2016 at 23:23
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    $\begingroup$ @zeraouliarafik, don't worry, in some sense this was a perfectly-answerable question for this site: it is certainly true that standard methods give the result, but asking to avoid analytic methods is a potentially interesting question about the combinatorics of cancellation/telescoping. Noam E. happened to recall at least the one prior occurrence, etc. Bingo. :) $\endgroup$
    – paul garrett
    Commented Jan 19, 2016 at 23:35
  • $\begingroup$ @paulgarrett, ok thank you very much , i will do an attempt to solve it $\endgroup$
    – zeraoulia rafik
    Commented Jan 19, 2016 at 23:36
  • $\begingroup$ @zeraouliarafik, but, of course, this might be very difficult, or in some sense impossible! :) $\endgroup$
    – paul garrett
    Commented Jan 19, 2016 at 23:38
  • $\begingroup$ amazing, that the 3 cannot get cancelled in a elementary manner. What do you make of that? $\endgroup$
    – Brad Graham
    Commented Mar 24, 2016 at 21:21
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Yes, as has been noted several times in comments, this has come up before, with a beautiful answer by David Speyer: Computing $\prod_p(\frac{p^2-1}{p^2+1})$ without the zeta function?

It seems to me this should put the present question to rest. The only "property" of the Riemann zeta function used is the Euler expansion $\sum_{n \geq 1} \frac1{n^s} = \prod_p \frac1{1 - p^{-s}}$, but the proof of this boils down to the fundamental theorem of arithmetic, which goes back to Euclid's Elements I think, so I'd hardly call this using (analytic) properties of the zeta function. Aside from that, David's demonstration (which is elementary in the technical sense) does the rest.

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  • $\begingroup$ See the comment thread for David Speyer's answer: he refrained from sending it in to the Monthly because a very similar argument for $\zeta(4) = (2/5) \zeta(2)^2$ had already been given by Zagier. $\endgroup$
    – Noam D. Elkies
    Commented Jan 20, 2016 at 16:34
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    $\begingroup$ @NoamD.Elkies Thanks; I did eventually see it but had no opportunity to edit until now. $\endgroup$
    – Todd Trimble
    Commented Jan 20, 2016 at 16:52

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