Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 150 / 1

Let $1(n)=1$. The basic Von Mangoldt divisor identity isbecause if $n={\prod }_p{p}^{v_p(n)}$, the right-hand side is ${\sum }_p{v}_p(n)\log p=\log n$. In terms of Dirichlet convolution, this says $\log =1\ast \Lambda$. Since the Möbius function is the convolution inverse of $1$, convolving with $\mu$ gives $\Lambda =\mu \ast \log$. Consequently

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The Dirichlet hyperbola method counts each factorization $n=ab$ once and gives, with $y=\lfloor \sqrt{x}\rfloor$,Using $\lfloor x/a\rfloor =x/a+O(1)$ and the harmonic number estimate $H_y=\log y+\gamma +O(1/y)$, where $\gamma$ is the Euler--Mascheroni constant, we obtainThus one may take $C=2\gamma -1$.

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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

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