 Solution

ID: past-exam-of-the-short-options-course-of-the-university-of-oxford/2017/1/c/solution

ϕ (n) = # {1 \leq a \leq n : g cd (a, n) = 1} .

n = \prod_{j = 1}^{r} p_{j}^{e_{j}}

with distinct prime numbers

p_{j}

ϕ (n) = n j = 1 \prod r (1 - \frac{1}{p _{j}}) = j = 1 \prod r p_{j}^{e_{j} - 1} (p_{j} - 1) .

In particular,

ϕ (2017) = 2016

. Adjoining the prime factor

2

changes neither the odd factor nor its count of units, since

ϕ (2) = 1

n = 4034 = 2 \cdot 2017, ϕ (4034) = (2 - 1) (2017 - 1) = 2016.

This is a positive integer distinct from

2017

, as required.

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