Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 117 / 1 / c

Table of contents
- Solution c

Solution

Put

F (m) = m (m + 2) (m + 6)

and let

ρ (d)

be the number of roots of

F

d

. For squarefree

d

, the Chinese remainder theorem gives

ρ (d) = \prod_{p ∣ d} ρ (p), ρ (2) = 1, q u a d ρ (3) = 2, q u a d ρ (p) = 3 (p \geq 5) .

Indeed, for

p \geq 5

the three roots

0, - 2, - 6

are distinct. Counting

m \leq X

in each of the

ρ (d)

residue classes gives

∣ A_{d} ∣ = # {m \leq X : d ∣ F (m)} = \frac{ρ ( d )}{d} X + O (ρ (d)) .

Consequently a suitable sieve distribution is

g (d) = \frac{ρ ( d )}{d}, r_{d} = O (ρ (d)) = O (3^{ω (d)}) .

This is the polynomial root density in a sieve calculation.

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