 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ii/paper-4/11i/a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

N

b

when it is composite, coprime to

b

, and

b^{N - 1} \equiv 1 (mod N)

. Here

N ∣ b^{k} - 1

gives coprimality and

k ∣ N - 1

, so

b^{N - 1} = (b^{k})^{(N - 1) / k} \equiv 1 (mod N)

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