Two Zetas in One

Relevant wiki: Riemann Zeta Function

According to Riemann Zeta function , the product of prime $p$ can be written in a form:

$\zeta(s)=\prod_{p=prime} \dfrac{1}{1-p^{-s}}.$

Then we need to rearrange the original expression into a more applicable form:

$\prod_{p=prime} (\dfrac{1}{1-p^{-1}} + \dfrac{1}{1+p^{-1}} - 1) = \prod_{p=prime} (\dfrac{p}{p-1} + \dfrac{p}{p+1} - 1) = \prod_{p=prime} (\dfrac{2p^2}{p^2 -1} -1) = \prod_{p=prime} (\dfrac{p^2 +1}{p^2 -1})$

Dividing the numerator and the denominator with $p^2$ :

$\prod_{p=prime} (\dfrac{p^2 +1}{p^2 -1}) = \prod_{p=prime} (\dfrac{1 + p^{-2}}{1- p^{-2}})$

Then multiplying both with $1-p^{-2}$ :

$\prod_{p=prime} (\dfrac{1 + p^{-2}}{1- p^{-2}}) = \prod_{p=prime} (\dfrac{1 - p^{-4}}{(1- p^{-2})^2})$

Hence, this is in a form of $\dfrac{(\zeta(2))^2}{\zeta(4)} = (\dfrac{\pi^2}{6})^2 \dfrac{90}{\pi^4} = \boxed{2.5}$

Unlike Worranat Pakornrat I had a sloppy beginning

$\prod _{p} \left ( \frac{1}{1-p^{-1}} + \frac{1}{1+p^{-1}} - 1 \right )$

$= \prod_{p} \left ( \sum_{n=0}^{\infty} \left ( \left(\frac{1}{p} \right )^{n} + \left (- \frac{1}{p}\right )^{n}\right)-1 \right )$

$= \prod_{p} \left ( 2\sum_{n=0}^{\infty} \left ( \frac{1}{p^{2n}} \right ) - 1\right)$

$= \prod_{p}\left( 2 \left( \frac{p^{2}}{p^{2} - 1} \right ) - 1 \right )$

$= \prod_{p} \left ( \frac{2p^{2}}{p^{2} - 1} - 1\right ) = \prod_{p} \left ( \frac{p^{2} + 1}{p^{2} - 1}\right)$

I then proceeded as Worranat. I am aware that I most likely made many mistakes along the way, regardless, this is my "reasoning".