Computing $\prod_p(\frac{p^2-1}{p^2+1})$ without the zeta function? 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 $\frac{2}{5}$: $\prod_p(\frac{p^2-1}{p^2+1})=\frac{3}{5} \cdot \frac{8}{10} \cdot \frac{24}{26} \cdot \frac{48}{50} \cdots$. 

A proof of this identity not using properties of the Riemann zeta function is listed as an unsolved problem in section B48 of Guy's Unsolved Problems in Number Theory.
An amusing observation: this identity implies the infinitude of primes. If the product over $p$ were finite, the final answer would be a rational number with a factor of $3$ in its numerator --- the very first term in the product has a $3$ in the numerator, and $3$ cannot divide $u^2+1$ for any integer $u$, so the $3$ can never cancel.
