Infinite prime product equals $\sqrt{5}$

Question

If $p$ runs through the primes of form $20k+3$ and $20k+7$ and $q$ runs through the primes of form $20k+13$ and $20k+17$, then $$\prod\limits^{\infty}\frac{p+1}{p-1}\prod\limits^{\infty}\frac{q-1}{q+1}=\sqrt{5}$$ How can I prove it?

Answer 1

The product

$$P(s)=\prod_{p\equiv3,7\bmod{20}}\frac{p^s+1}{p^{s}-1} \prod_{p\equiv13,17\bmod{20}}\frac{p^s-1}{p^{s}+1}$$

can be written, for $\Re(s)>1$, as a quotient of two Dirichlet $L$-series:

$$P(s)=\frac{L(s,\chi_1)}{L(s,\chi_2)}$$

where $\chi_1(k+20\mathbf Z)=-1$ iff $k=11,13,17,19$ and $\chi_1(k+20\mathbf Z)=1$ otherwise, and $\chi_2(k+20\mathbf Z)=-1$ iff $k=3,7,11,19$ and $\chi_2(k+20\mathbf Z)=1$ otherwise (of course, when $(k,n)>1$, both are zero). For $s=1$ it can be shown that

$$\frac{\overbrace{L(1,\chi_1)}^{\pi/\sqrt5}}{\underbrace{L(1,\chi_2)}_{\pi/5}}=\sqrt5$$

(the calculations are cumbersome, I've used Mathematica). Thus, we might expect the equality

$$P(1)"="\sqrt 5$$

but we are not allowed to do that, since the equality is valid only on the half-plane $\Re(s)>1$. Maybe someone can justify this last step, by rearranging the product in an appropriate manner.

Answer 2

As i started the answer, the quick, compact and good answer (+1 from me) of iqcd was not here, so i switched to an editor, but typing killed me. The answer was here in between, same and only possible idea, will post it, since some details may bring a new information.

Let $T$ be the group of units of the ring $\Bbb Z/20\Bbb Z$. It has $\phi(20)=20\cdot\left(1-\frac 12\right)\left(1-\frac 15\right)=8$ elements, and it is isomorphic to... $$ T\cong(\Bbb Z/4)\times(\Bbb Z/2)\ .$$

For the sake of typography, i will use sage to have a quick information, here and in the sequel.

sage: R = ZZ.quotient(20)
sage: R
Ring of integers modulo 20
sage: G = R.unit_group()
sage: G.order()
8
sage: G
Multiplicative Abelian group isomorphic to C2 x C4

As generators, we take $11$, with $11^2=121=1$ (in $T$), and $17$, with powers $1,17,(28)9, (49)13$ (in $T$).

The characters are as in the following listing:

sage: D = DirichletGroup(20)
sage: for chi in D:
....:     print chi
....:     
Dirichlet character modulo 20 of conductor 1 mapping 11 |--> 1, 17 |--> 1
Dirichlet character modulo 20 of conductor 4 mapping 11 |--> -1, 17 |--> 1
Dirichlet character modulo 20 of conductor 5 mapping 11 |--> 1, 17 |--> zeta4
Dirichlet character modulo 20 of conductor 20 mapping 11 |--> -1, 17 |--> zeta4
Dirichlet character modulo 20 of conductor 5 mapping 11 |--> 1, 17 |--> -1
Dirichlet character modulo 20 of conductor 20 mapping 11 |--> -1, 17 |--> -1
Dirichlet character modulo 20 of conductor 5 mapping 11 |--> 1, 17 |--> -zeta4
Dirichlet character modulo 20 of conductor 20 mapping 11 |--> -1, 17 |--> -zeta4

and we want to see the characters from above, calculated at all places.

Here they are. Code:

D = DirichletGroup(20)
V = ( 1, 3, 7, 9, 11, 13, 17, 19 )
for v in V:
    print "%6s" % v,
print 

for chi in D:
    for v in V:
        print "%6s" % chi(v),
    print

Results:

 1      3      7      9     11     13     17     19
 1      1      1      1      1      1      1      1
 1     -1     -1      1     -1      1      1     -1
 1 -zeta4  zeta4     -1      1 -zeta4  zeta4     -1
 1  zeta4 -zeta4     -1     -1 -zeta4  zeta4      1
 1     -1     -1      1      1     -1     -1      1
 1      1      1      1     -1     -1     -1     -1
 1  zeta4 -zeta4     -1      1  zeta4 -zeta4     -1
 1 -zeta4  zeta4     -1     -1  zeta4 -zeta4      1

We will now type only the two lines that are mostly closed to the given problem:

 1      3      7      9     11     13     17     19
 1     -1     -1      1     -1      1      1     -1
 1      1      1      1     -1     -1     -1     -1

and the connection to the problem is as follows. Consider the Dirichlet $L$-series associated to a Dirichlet character of $\Bbb Z/20$, in usual notation $\chi$, Dirichlet $L$-function by $$ L(s,\chi) =\sum_{n\ge 1}\frac {\chi(n)}{n^s} =\prod_{p\text{ prime}}\left(1-\frac {\chi(p)}{p^s}\right)^{-1}\ . $$ The above Euler product is the one we are interested in.

The quotient of the two $L$-functions, associated to the two characters above is the value.

In our case, formally: $$ \begin{aligned} &\prod_{\substack{p\equiv 3,7[20]\\p\text{ prime}}}\frac {p+1}{p-1} \prod_{\substack{q\equiv 13,17[20]\\q\text{ prime}}}\frac {q-1}{q+1} \\ &\qquad= \prod_{\substack{p\equiv 3,7[20]\\p\text{ prime}}}\frac {1+\frac 1p}{1-\frac 1p} \prod_{\substack{q\equiv 13,17[20]\\q\text{ prime}}}\frac {1-\frac 1q}{1+\frac 1q} \\ &\qquad= \frac {\prod_{p\text{ prime}}\left(1-\frac {\chi_1(p)}{p}\right)} {\prod_{p\text{ prime}}\left(1-\frac {\chi_2(p)}{p}\right)} \\ &\qquad= \frac {\prod_{p\text{ prime}}\left(1-\frac {\chi_2(p)}{p}\right)^{-1}} {\prod_{p\text{ prime}}\left(1-\frac {\chi_1(p)}{p}\right)^{-1}} \\ &\qquad= \frac {L(\chi_2,1)} {L(\chi_1,1)} \\ &\qquad= \frac {\pi/\sqrt 5} {\pi/5} =\sqrt 5 \ . \end{aligned} $$ With the "obvious" choices for $\chi_1,\chi_2$ from the above table. Here, $$ L(1,\chi) = \frac {\pi i}{20} \sum_{1\le a\le 9}\chi(a)\frac {\zeta^a+1}{\zeta^a-1}\ , $$ where $\zeta$ is the primitive root of order $20$, $\zeta=\exp\frac{2\pi i}{20}$. In our cases, we have used above:

sage: def L(chi):    return pi*i/20* sum( [chi(a) * (z^a+1)/(z^a-1) for a in [1..9]] )
sage: L( G.list()[1] ).simplify_full()
1/5*pi
sage: L( G.list()[5] ).simplify_full()
1/5*sqrt(5)*pi

(Sage was used again.)

Note: There are some issues on convergence.

The right hand side of $$ L(s,\chi) =\sum_{n\ge 1}\frac {\chi(n)}{n^s} %=\prod_{p\text{ prime}}\left(1-\frac {\chi(p)}{p^s}\right)^{-1} $$ is absolutely convergent for $s$ with real part $>1$, and convergent for a non-trivial character, as in our cases, for $s$ with real part $>0$.

This is still not enough for using it, its partial sums, instead of partial Euler products in the formal computation above. The truncation of the products have to be controlled somehow. (I will try to look closer...)