Prime Product

Number Theory Level 5

It is a well known result that

∏ p is prime 1 − p − 2 1 + p − 2 = 2 5 \prod _{p \text{ is prime}}\dfrac{1-p^{-2}}{1+p^{-2}}=\dfrac{2}{5}

Now, if

∏ p ≡ − 1 ( m o d 4 ) 1 − p − 2 1 + p − 2 = A G π B , \displaystyle \large \prod _{ p\equiv-1\pmod4 }\dfrac{1-p^{-2}}{1+p^{-2}}=\frac{AG}{\pi^B},

where A A and B B are integers with G G denotes the Catalan's constant . Find A + B A+B .


The answer is 10.

Julian Poon by Julian Poon , 20, Singapore

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Otto Bretscher
Apr 23, 2016

This is a beautiful problem of elegant simplicity! Thanks!

Consider the Dirichlet Beta Function β ( s ) = ∑ n = 0 ∞ ( − 1 ) n ( 2 n + 1 ) s \beta(s)=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^s} whose Euler Product is β ( s ) = ∏ p ≡ 1 1 1 − p − s ∏ p ≡ 3 1 1 + p − s \beta(s)=\prod_{p\equiv 1}\frac{1}{1-p^{-s}}\prod_{p\equiv 3}\frac{1}{1+p^{-s}} where congrences are taken modulo 4. Now ζ ( s ) = 1 1 − 2 − s ∏ p ≡ 1 1 1 − p − s ∏ p ≡ 3 1 1 − p − s \zeta(s)=\frac{1}{1-2^{-s}}\prod_{p\equiv 1}\frac{1}{1-p^{-s}}\prod_{p\equiv 3}\frac{1}{1-p^{-s}} so that ∏ p ≡ 3 1 − p − s 1 + p − s = β ( s ) ζ ( s ) ( 1 − 2 − s ) \prod_{p\equiv 3}\frac{1-p^{-s}}{1+p^{-s}}=\frac{\beta(s)}{\zeta(s)(1-2^{-s})} .

For s = 2 s=2 this is 8 G π 2 \frac{8G}{\pi^2} , and the answer is 10 \boxed{10} ; recall that the Catalan constant is defined to be β ( 2 ) \beta(2) .

4 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Oh wow, that's a nice way to interpret this problem. Thanks for sharing the approach :)

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...