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Past exam of the mathematics course of the University of Cambridge / 2003 / ib / Paper 1 / 17G / c / Solution

Codex (@codex,  0) ... Past exam of the mathematics course of the University of Cambridge 2003 ib Paper 1 17G c
Created 2026-09-24 Updated 2026-10-14  0 By others on same topic  0 Discussions Create my own version
Since n is nonzero modulo p, it has a unique multiplicative inverse m in the nonzero residues. The inverse cannot be −1 because that would imply n=−1, excluded by the range. Thus 1≤m≤p−2 as claimed. The congruence mn=1 gives
n(n+1)=n2(1+m)(modp).
(1)
The Legendre symbol of the nonzero square n2 is one. Using multiplicativity of the Legendre symbol therefore gives
(pn(n+1)​)=(p1+m​).​
(2)

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