 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ib/paper-2/1e/i/solution

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

Let

P_{n} (X) = 1 + X + \dots + X^{n - 1} .

n = ab

with

a, b > 1

, then

P_{n} (X) = (1 + X + \dots + X^{a - 1}) (1 + X^{a} + \dots + X^{a (b - 1)}),

P_{n}

is reducible.

Conversely, let

n = p

be prime. Translation by one is an automorphism of

Z [X]

, and

P_{p} (X + 1) = \frac{( X + 1 ) ^{p} - 1}{X} = \sum_{k = 1}^{p} (k p) X^{k - 1} .

Its leading coefficient is one, every lower coefficient is divisible by

p

, and its constant coefficient is

p

, which is not divisible by

p^{2}

P_{p} (X + 1)

is irreducible, hence so is

P_{p} (X)

1 + X + \dots + X^{n - 1} is irreducible in Z [X] ⟺ n is prime .

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