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Prove that for any prime $p>3$, the number $\binom{2p-1}{p-1}-1$ is divisible by $p^{3}$.


Since every integer that is relatively prime to p has a multiplicative inverse modulo p, denote the inverse of x modulo p by x−1. We start by improving the conclusion of the previous problem. Therefore: $$\binom{2p}{p}-2=\sum_{k=1}^{p-1}\binom{p}{k}^{2}=\sum_{k=1}^{p-1}\left ( \frac{p}{k}\binom{p-1}{k-1} \right )^{2}$$

I feel like I have made a mistake because unless $\frac{1}{k}\binom{p-1}{k-1}$ is an integer I can't proceed any further. Is it an integer? Where should I go from here?

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1 Answer 1

The expression eventually turns out to be $\dfrac{1}{2}$ $ 2p \choose p$$-1$ from here .

Wolstenholme's theorem:

$ \dfrac{1}{2}{2p\choose p}\equiv 1\mod p^3$

Just Collecting the facts.

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