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 Incoming links: Weighted complete Legendre-symbol sum

Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 4 / 11G / c / ii / Solution Created 2026-09-24 Updated 2026-09-25

By Euler criterion,

T \equiv \sum_{a = 1}^{p - 1} a^{(p + 1) /2} (mod p) .

Choose a primitive root

g

. For any integer

m

not divisible by

p - 1

,

\sum_{a = 1}^{p - 1} a^{m} = \sum_{j = 0}^{p - 2} g^{jm} = \frac{g ^{m (p - 1)} - 1}{g ^{m} - 1} \equiv 0 (mod p) .

For

p > 3

, the exponent

m = (p + 1) /2

satisfies

0 < m < p - 1

. Therefore

T \equiv 0 (mod p) .

Together with part (i), this is the weighted complete Legendre-symbol sum.

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