Is it possible to derive the following $\zeta$ relations without actually finding the values themselves?
\begin{align*} 2 \zeta(2)^2 &= 5\zeta(4)\\ 4\zeta(2)\zeta(4) &= 7\zeta(6)\\ 3\zeta(2)\zeta(6) &= 5\zeta(8) \end{align*}
These small integer relations make it look like there is a nice relation between these values.
However, the pattern breaks at the next one and the simplest relation I can find is $4+6=10$
$$10\zeta(4)\zeta(6) = 11\zeta(10)$$
I thought of using the integral representation of $\zeta(s)$ but I do not see an obvious way forward.
$\mathbf{\text{Hint:}}$(after seeing that you are much interested for even zeta values)
The pattern you look towards is seen from the fact that $$\zeta(2n)=\frac{(-1)^{n+1}(2\pi)^{2n}B_{2n}}{2(2n)!}$$ Where $B_n$ are the Bernoulli numbers and $n\in\Bbb{N}$