Past exam of the mathematics course of the University of Cambridge / 2019 / ia / Paper 4 / 6E / d

Table of contents
- Solution d

Solution

The class of

(1, 1)

is

{(λ, λ) : λ \in (Z / n Z)^{\times}},

so its size is Euler's totient function

φ (n) = n \prod_{p ∣ n} (1 - \frac{1}{p}) .

The unit group acts freely on every pair in

Y

: if

λa \equiv a

and

λb \equiv b

, multiplying the Bezout identity by

λ - 1

gives

λ \equiv 1 (mod n)

. Hence every equivalence class has size

φ (n)

. Dividing the answer from part (a) by this size gives

\frac{∣ Y ∣}{φ ( n )} = n p ∣ n \prod \frac{1 - p ^{- 2}}{1 - p ^{- 1}} = n p ∣ n \prod (1 + \frac{1}{p}) .

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