Showing irrationality of $\zeta(k)$ for some $k$ without calculating the value. For $s\in (1,\infty)$ let $\zeta(s):=\sum_{n=1}^\infty \dfrac 1{n^s}$. 
Is there a way to show that $\zeta(2k)$ is irrational for some integer $k\geq 1$ without finding explicit formulae? 
 A: I can prove $\zeta(4)$ is irrational without needing to calculate it.
Consider for $z$ in the upper half plane the weight $4$ Eisenstein series:
$G_4(z) = \sum_{(m,n)\in\mathbb{Z}^2\backslash (0,0)}\frac{1}{(mz+n)^4}$
It has a Fourier expansion:
$G_4(z) = 2\zeta(4) + \frac{16\pi^4}{3}\sum_{n=1}^{\infty}\sigma_{3}(n) q^n$
where $q = e^{2\pi i z}$ and $\sigma_3(n) = \sum_{d\mid n}d^3$. Assuming we know that $\pi^4$ is irrational (which follows from the fact that $\pi$ is transcendental) then it is clear that all Fourier coefficients for $n\geq 1$ are irrational. 

Also consider the theta series $\theta_{E_8}(z)$ for the $E_8$ lattice. This has a Fourier expansion:
$\theta_{E_8}(z) = \sum_{n=0}^{\infty}r(n) q^n$
where the $r(n)$ values are all integers (they measure numbers of vectors in $E_8$ of a given norm).

Here is the magic, both $G_4, \theta_{E_8}$ are modular forms of weight $4$ for the modular group SL$_2(\mathbb{Z})$.
However it is known that the $\mathbb{C}$-vector space of such functions is $1$-dimensional. Thus $\theta_{E_8} = \alpha G_4$ for some $\alpha\in\mathbb{C}^{\times}$. So in particular there is a number $\alpha$ that scales all Fourier coefficients of $G_4$ to be integers.
But if we suppose $\zeta(4)\in\mathbb{Q}$ then we hit a problem since then the Fourier coefficients of $G_4$ would be a mixture of rational and irrational and no such scaling could then simultaneously produce integers (if $x$ is rational and $y$ is irrational then at most one of $\alpha x, \alpha y$ can be rational for any $\alpha\in\mathbb{C}^{\times}$). 
Actually it is easy to get the usual formula $\zeta(4) = \frac{\pi^4}{90}$ from the equality $\theta_{E_8}(z) = \alpha G_4$.
