 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ia/paper-4/6e/g/solution

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

Choose a prime with

p^{2} ∣ n

, and set

a = 0

b = n / p

. Then

b \neq \equiv 0 (mod n)

, but

b^{2} = n^{2} / p^{2}

is divisible by

n

. Therefore

a^{k} \equiv b^{k} \equiv 0 (mod n)

for every

k > 1

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