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Divisibility of a sum by a prime p

Source: Tuymaada 2012, Problem 4, Day 1, Seniors

July 20, 2012
quadraticsmodular arithmeticcalculusalgebrapolynomial

Problem Statement

Let p=4k+3p=4k+3 be a prime. Prove that if 102+1+112+1+⋯+1(p−1)2+1=mn\dfrac {1} {0^2+1}+\dfrac{1}{1^2+1}+\cdots+\dfrac{1}{(p-1)^2+1}=\dfrac{m} {n} (where the fraction mn\dfrac {m} {n} is in reduced terms), then p∣2m−np \mid 2m-n.
Proposed by A. Golovanov
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