 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ii/paper-4/11h/d/solution

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

n = \prod_{p^{α} ∥ n} p^{α}

, multiplicativity and the prime-power formula for the divisor function give

\frac{τ ( n )}{n ^{ϵ}} = \prod_{p^{α} ∥ n} \frac{α + 1}{p ^{α ϵ}} .

Choose

P_{0} = 2^{1/ ϵ}

. For

p \geq P_{0}

and

α \geq 1

, the elementary bound

α + 1 \leq 2^{α}

gives

\frac{α + 1}{p ^{α ϵ}} \leq 1.

For each of the finitely many primes

p < P_{0}

(α + 1) p^{- α ϵ}

tends to zero and therefore has a finite maximum

M_{p}

over

α \geq 0

. Consequently

\frac{τ ( n )}{n ^{ϵ}} \leq C (ϵ) := \prod_{p < P_{0}} M_{p} < \infty,

uniformly in

n

, which proves

τ (n) \leq C (ϵ) n^{ϵ}

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