We see that $\frac{2}{5}=\frac{36}{90}=\frac{6^2}{90}=\frac{\zeta(4)}{\zeta(2)^2}=\Pi_p\frac{(1-\frac{1}{p^2})^2}{(1-\frac{1}{p^4})}=\Pi_p(\frac{(p^2-1)^2}{(p^2+1)(p^2-1)})=\Pi_p(\frac{p^2-1}{p^2+1})$

$\implies \Pi_p(\frac{p^2-1}{p^2+1})=\frac{2}{5}$.

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 $\Pi_p(\frac{p^2-1}{p^2+1})$ also admits the factorization $\Pi_p(\frac{p-1}{p-i})\Pi_p(\frac{p+1}{p+i})$. Also notice that numerically it is quite obvious that the product is convergent to $\frac{2}{5}$: $\Pi_p(\frac{p^2-1}{p^2+1})=\frac{3}{5} \cdot \frac{8}{10} \cdot \frac{24}{26} \cdot \frac{48}{50} \cdot \dots$.

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

We see that $\frac{\zeta(4)}{\zeta(2)^2}=\frac{6^2}{90}=\frac{36}{90}=\frac{2}{5}=\Pi_p\frac{(1-\frac{1}{p^2})^2}{(1-\frac{1}{p^4})}=\Pi_p(\frac{(p^2-1)^2}{p^2+1})=\Pi_p(\frac{p^2-1}{p^2+1})$

$\implies \Pi_p(\frac{p^2-1}{p^2+1})=\frac{2}{5}$.

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 $\Pi_p(\frac{p^2-1}{p^2+1})$ also admits the factorization $\Pi_p(\frac{p-1}{p-i})\Pi_p(\frac{p+1}{p+i})$. Also notice that numerically it is quite obvious that the product is convergent to $\frac{2}{5}$: $\Pi_p(\frac{p^2-1}{p^2+1})=\frac{3}{5} \cdot \frac{8}{10} \cdot \frac{24}{26} \cdot \frac{48}{50} \cdot \dots$.

We see that $\frac{2}{5}=\frac{36}{90}=\frac{6^2}{90}=\frac{\zeta(4)}{\zeta(2)^2}=\Pi_p\frac{(1-\frac{1}{p^2})^2}{(1-\frac{1}{p^4})}=\Pi_p(\frac{(p^2-1)^2}{(p^2+1)(p^2-1)})=\Pi_p(\frac{p^2-1}{p^2+1})$

$\implies \Pi_p(\frac{p^2-1}{p^2+1})=\frac{2}{5}$.

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 $\Pi_p(\frac{p^2-1}{p^2+1})$ also admits the factorization $\Pi_p(\frac{p-1}{p-i})\Pi_p(\frac{p+1}{p+i})$. Also notice that numerically it is quite obvious that the product is convergent to $\frac{2}{5}$: $\Pi_p(\frac{p^2-1}{p^2+1})=\frac{3}{5} \cdot \frac{8}{10} \cdot \frac{24}{26} \cdot \frac{48}{50} \cdot \dots$.

We see that $\frac{\zeta(4)}{\zeta(2)^2}=\frac{6^2}{90}=\frac{36}{90}=\frac{2}{5}=\Pi_p\frac{(1-\frac{1}{p^2})^2}{(1-\frac{1}{p^4})}=\Pi_p(\frac{(p^2-1)^2}{p^2+1})=\Pi_p(\frac{p^2-1}{p^2+1})$

$\implies \Pi_p(\frac{p^2-1}{p^2+1})=\frac{2}{5}$.

But is this the only way to compute this product of infinite primes? It seems like such a simple product, one that could be calculated without the zeta function.

Note that $\Pi_p(\frac{p^2-1}{p^2+1})$ also admits the factorization $\Pi_p(\frac{p-1}{p-i})\Pi_p(\frac{p+1}{p+i})$. Also notice that numerically it is quite obvious that the product is convergent to $\frac{2}{5}$: $\Pi_p(\frac{p^2-1}{p^2+1})=\frac{3}{5} \cdot \frac{8}{10} \cdot \frac{24}{26} \cdot \frac{48}{50} \cdot \dots$.

We see that $\frac{\zeta(4)}{\zeta(2)^2}=\frac{6^2}{90}=\frac{36}{90}=\frac{2}{5}=\Pi_p\frac{(1-\frac{1}{p^2})^2}{(1-\frac{1}{p^4})}=\Pi_p(\frac{(p^2-1)^2}{p^2+1})=\Pi_p(\frac{p^2-1}{p^2+1})$

$\implies \Pi_p(\frac{p^2-1}{p^2+1})=\frac{2}{5}$.

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 $\Pi_p(\frac{p^2-1}{p^2+1})$ also admits the factorization $\Pi_p(\frac{p-1}{p-i})\Pi_p(\frac{p+1}{p+i})$. Also notice that numerically it is quite obvious that the product is convergent to $\frac{2}{5}$: $\Pi_p(\frac{p^2-1}{p^2+1})=\frac{3}{5} \cdot \frac{8}{10} \cdot \frac{24}{26} \cdot \frac{48}{50} \cdot \dots$.