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(1 + a^1)(1 + a^2)...(1 + a^{p - 1})\equiv 1 (mod q), if a^p \equiv 1 mod q

Source: 2020 Estonia TST 4.3

November 18, 2020
number theoryremainder

Problem Statement

The prime numbers pp and qq and the integer aa are chosen such that p>2p> 2 and a≢1a \not\equiv 1 (mod qq), but ap≡1a^p \equiv 1 (mod qq). Prove that (1+a1)(1+a2)...(1+ap−1)≡1(1 + a^1)(1 + a^2)...(1 + a^{p - 1})\equiv 1 (mod qq) .
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