olymmath

ELMO SL 2014 N6

    Show that the numerator of 2p−1p+1−(∑k=0p−1(p−1k)(1−kp)2)\frac{2^{p-1}}{p+1} - \left(\sum_{k = 0}^{p-1}\frac{\binom{p-1}{k}}{(1-kp)^2}\right) is a multiple of p3p^3 for any odd prime pp .

    Proposed by Yang Liu

    Dataset text AI-MO AoPS. Quoted with source.

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