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Apply part a to $A=\{1,\dots ,\lfloor x\rfloor \}$ and all primes. Sincewe obtainFor $z\le \log x$, the error is at most $2^z\le x^{\log 2}=o(x/\log z)$. By Mertens theorem, as $z\to \infty$,Thusso one may take $C=e^{-\gamma }>0$. For bounded $z$, the preceding exact product formula gives the corresponding fixed density.

Solved by gpt-5.6-sol high.

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.

Complete multiplicativity and absolute convergence give the Euler productSet $\sigma =1+1/\log x$. Taking logarithms of absolute values and expanding the local factors gives, uniformly in real $t$,The prime powers with exponent at least two contribute $O(1)$; changing $p^{-\sigma }$ to $p^{-1}$ below $x$ and estimating the tail above $x$ also cost $O(1)$. By Mertens theorem,Exponentiating yields

Solved by gpt-5.6-sol high.

For each prime number $p\nmid a$, the congruence $an+1\equiv 0(\mathrm{mod}p)$ removes exactly one residue class of $n$ modulo $p$; for $p\mid a$, it removes none. The dimension-one upper-bound sieve therefore givesSeparating the primes that divide $a$ bounds the product byThe ratio form of Mertens theorem says that the first product is $\ll \log w/\log z$. Hence

Solved by gpt-5.6-sol high.