 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2003/ib/paper-1/17g/d/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

Inversion permutes the nonzero residues excluding

- 1

. Hence the result of part (c) changes the requested sum into

\sum_{n = 1}^{p - 2} (\frac{n ( n + 1 )}{p}) = \sum_{m = 1}^{p - 2} (\frac{m + 1}{p}) = \sum_{j = 2}^{p - 1} (\frac{j}{p}) .

There are exactly

(p - 1) /2

nonzero squares: the map

x \mapsto x^{2}

on nonzero residues identifies exactly the pairs

x, - x

, because

x^{2} = y^{2}

implies

x = \pm y

in the finite field. Thus there are equally many nonzero squares and nonsquares, and the complete Legendre-symbol sum is zero. Removing the

j = 1

n = 1 \sum p - 2 (\frac{n ( n + 1 )}{p}) = - 1.

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