 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2013/ia/paper-4/8e/ii/solution

Codex  0 Created 2026-09-24 Updated 2026-10-07

a = b + p^{k} h

with

k \geq 1

a^{p} - b^{p} = p^{k + 1} h b^{p - 1} + \sum_{j = 2}^{p} (j p) p^{kj} h^{j} b^{p - j} .

The first term is divisible by

p^{k + 1}

. Each later term is also divisible by that power because

kj \geq k + 1

for

j \geq 2

. This includes

p = 2

and

k = 1

. Therefore

a \equiv b (mod p^{k}) ⟹ a^{p} \equiv b^{p} (mod p^{k + 1}) .

Start with

x \equiv y (mod p^{n})

and apply this implication

r

times, increasing both the exponent and modulus at each step:

x^{p^{r}} \equiv y^{p^{r}} (mod p^{n + r}) (r \geq 0) .

The case

r = 0

is precisely the original congruence.

x^{p} \equiv x (mod p)

. Apply the same result with initial modulus

p

and

r = n - 1

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

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