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

An elementary proof of the identity
$$ 2 \zeta(2)^2 = 5\zeta(4)$$
has been found by Don Zagier in the paper 
http://people.mpim-bonn.mpg.de/zagier/files/tex/ConsequencesCohomologySL/fulltext.pdf
The idea seems similar to how David Speyer proves it in that he starts by defining the function
$$ f(m,n) = \frac{2}{n^3m} + \frac{1}{n^2m^2} + \frac{2}{nm^3} $$
and then verifies that 
$$  f(m,n) -f(m,n+m) - f(m+n,n)  = \frac{2}{m^2n^2}.$$
But since we also have 
$$   \sum_{n,m>0} f(m,n)  - \sum_{n,m>0} f(m,n+m) - \sum_{m,n>0} f(m+n,n) = \sum_{n>0} f(n,n),$$
since only the diagonal terms survive, this gives the identity 
$$ 2 \zeta(2)^2 = 5\zeta(4).$$ 
The crucial function $f(m,n)$ seems to be part of a larger family of functions with ties to period polynomials, as explained in the paper. 
