2 added 44 characters in body; edited title

# A binomial sum is 1moddivisibleby p^2

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.

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# A binomial sum is 1 mod p^2

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.