 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ib/paper-3/11g/c/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

x^{2} - x + 1 = (x + ω) (x + ω^{2}) = y^{3} .

Let

π = 1 - ω

. Its norm is

3

, and

π^{2} = - 3 ω,

3

is a unit times

π^{2}

. Suppose

x \equiv 2 (mod 3)

, say

x = 3 k + 2

. Since

ω = 1 - π

x + ω = 3 (k + 1) - π .

The first term is divisible by

π^{2}

and the second has discrete valuation exactly one, so

v_{π} (x + ω) = 1

v_{π} (x + ω^{2}) = 1

as well. Consequently

v_{π} (x^{2} - x + 1) = 2.

But

v_{π} (y^{3}) = 3 v_{π} (y)

is divisible by three, a contradiction. Hence

x \neq \equiv 2 (mod 3) .

 New to topics? Read the docs here!