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Inverse modulo p!

Source: India TST 2001 Day 4 Problem 2

January 31, 2015
modular arithmeticnumber theory proposednumber theory

Problem Statement

Let p>3p > 3 be a prime. For each k∈{1,2,…,p−1}k\in \{1,2, \ldots , p-1\}, define xkx_k to be the unique integer in {1,…,p−1}\{1, \ldots, p-1\} such that kxk≡1(modp)kx_k\equiv 1 \pmod{p} and set kxk=1+pnkkx_k = 1+ pn_k. Prove that : ∑k=1p−1knk≡p−12(modp)\sum_{k=1}^{p-1}kn_k \equiv \frac{p-1}{2} \pmod{p}
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