Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 150 / 1 / c

LetFor every prime $p>5$, the congruence $F(n)\equiv 0(\mathrm{mod}p)$ excludes the three distinct residue classesThe finitely many smaller primes only alter the implied constant. The dimension-three upper-bound sieve, used with $z=x^{1/2}$, therefore giveswhere the middle estimate follows from Mertens theorem.

If all three linear forms are prime, then either $F(n)$ has no prime divisor at most $z$, or one of the three forms itself equals such a prime. The latter possibility contributes only $O(\pi (z))=O(x^{1/2})$, which is absorbed by $x/(\log x{)}^3$. Hence the required number of $n$ is $\ll x/(\log x{)}^3$.

Solved by gpt-5.6-sol high.