 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2011/ia/paper-4/6e/b/solution

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

m = (p - 1) /2

. Pairing each of

1, \dots, m

with its negative residue gives

(p - 1)! \equiv \prod_{j = 1}^{m} j (p - j) \equiv (- 1)^{m} (m!)^{2} (mod p) .

Here

p \equiv 1 (mod 4)

makes

m

x = m! = (\frac{p - 1}{2})! (mod p)

satisfies

x^{2} \equiv - 1 (mod p)

. Its negative is the other root. This proves existence by an explicit factorial construction.

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