This is a question I have since longer time, but I have absolutely no idea how to proceed on it.


Let $p>3$ be a prime. Prove that $\displaystyle\sum\limits\_{k=1}^{p-1}\frac{1}{k}\binom{2k}{k}\equiv 0\mod p^2$. Here, we work in $\mathbb{Z}\_{\mathbb{Z}\setminus p\mathbb{Z}}$ (that is, $\mathbb{Z}$ localized at all numbers not divisible by $p$).


I know that it is $0\mod p$ (though I can't find the reference at the moment; it was some hard olympiad problem on MathLinks). The $0\mod p^2$ assertion is backed up by computation for all $p<100$. I am sorry if this is trivial or known. I would be delighted to see a combinatorial proof (= finding a binomial identity which reduces to the above when computed $\mod p^2$). Some number-theoretical arguments would be nice, too. However, I fear that if you use analytic number theory, I will not understand a single word.


EDIT: Epic fail at question title fixed.



A binomial sum is divisible by p^2
Your binomial sum is divisible by $p^2$ as is shown in a recent paper of Sun and Tauraso:
<https://arxiv.org/abs/0805.0563>


Even more remarkably, they compute that it's equivalent to $\frac{8}{9}p^2B\_{p-3} \bmod p^3$, as well as various other congruences for the corresponding alternating sum.


