Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 150 (Analytic Number Theory)

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

Solved by gpt-5.6-sol high.

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 $C=2\gamma -1$.

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.

If $f$ is a multiplicative function with $|f(n)|\le 1$, then for $\Re s>1$ its Dirichlet series has the Euler productIndeed, expanding the product over a finite set of primes and using unique prime factorization gives the sum over integers having no other prime factors. Moreover,so absolute convergence permits rearrangement and passage to the limit over all primes. This proves the formula.

Solved by gpt-5.6-sol high.

The symmetric form of the Functional equation of the Riemann zeta function isThe complex conjugation identity $\zeta (\overline{s})=\overline{\zeta (s)}$ and the functional equation show that every Nontrivial zero of the Riemann zeta function $\rho$ is accompanied by $\overline{\rho }$, $1-\rho$, and $1-\overline{\rho }$.

Let $\rho =\beta +i\gamma$ be the given nonreal zero. If $\beta >1/2$, use $\rho$ itself; if $\beta <1/2$, use $1-\rho$, whose real part is $1-\beta >1/2$. The resulting zero cannot have real part greater than one, by the stated zero-free half-plane. It therefore has real part in $(1/2,1]$.

Solved by gpt-5.6-sol high.

Write $M(x)={\sum }_{n\le x}\mu (n)$ for the Mertens function. Suppose, to the contrary, that for some $\epsilon >0$ the quotient $|M(x)|/x^{1/2-\epsilon }$ were bounded. Partial summation would then makeconverge and define a holomorphic function throughout $\Re s>1/2-\epsilon$. In $\Re s>1$ the Euler product identifies this function with $1/\zeta (s)$, so analytic continuation would make $1/\zeta (s)$ holomorphic in that larger half-plane.

By assumption, $\zeta$ has a nontrivial zero. The Functional equation of the Riemann zeta function reflects one of that zero and its partner into $\Re s\ge 1/2$, where $1/\zeta$ must have a pole, a contradiction. Thus $|M(x)|/x^{1/2-\epsilon }$ is unbounded, which gives an $x\ge 1$ exceeding any prescribed constant $C$.

Solved by gpt-5.6-sol high.

One quantitative form of Halász theorem is the following. If $f$ is multiplicative and $|f(n)|\le 1$, putThen, uniformly for $2\le T\le x$,Thus a bounded multiplicative arithmetic function can have a large mean only when it has small pretentious distance from some Archimedean character $n^{it}$.

Solved by gpt-5.6-sol high.

Put $g=f\mu$. At every prime, $f(p{)}^2=1$, so the triangle inequality for pretentious distance givesThe standard strong aperiodicity of the Möbius function states, for example with $T=(\log x{)}^{1/10}$, thatIndeed, its left side is controlled by the prime sum ${\sum }_{p\le x}(1+\cos (v\log p))/p$, uniformly in this range.

Choose $t$ and $u$ minimizing the two distances in Halász theorem. The displayed triangle inequality implies that at least one of $M(f;x,T)$ and $M(g;x,T)$ tends to infinity. Halász's bound, and $T\to \infty$, then show that at least one oftends to zero. Their minimum is consequently $o(1)$, which is the claimed $o(x)$ estimate before normalization.

Solved by gpt-5.6-sol high.

Let $F(s)={\sum }_{n\ge 1}{a}_n{n}^{-s}$ converge absolutely for $\Re s>{\sigma }_a$. Perron formula states that for $c>{\sigma }_a$ and nonintegral $x>0$,where the integral is understood as the limit of symmetric truncations under the usual convergence hypotheses. If $x$ is an integer, the endpoint term is counted with weight $1/2$. Effective versions truncate at height $T$ and include an explicit error depending on the coefficients and the distance of $x$ from nearby integers.

Solved by gpt-5.6-sol high.

Put $t=s-1$. The Laurent series of the logarithmic derivative at the simple pole of $\zeta$ has the formwhere in fact $c_0=\gamma$. HenceThe coefficient of $t^{-1}$ in their product, and therefore the residue, iswith the constant $A=-1-2c_0$; equivalently, $A=-1-2\gamma$.

Solved by gpt-5.6-sol high.

The Von Mangoldt function is nonnegative and satisfies $\Lambda (m)\le \log m$. Using the Von Mangoldt divisor identity,

For $0<u\le 1$, a comparison with an improper integral gives

Solved by gpt-5.6-sol high.

For $\Re s>1$, the Dirichlet series multiplication rule and $-{\zeta }^{\prime }/\zeta (s)={\sum }_{n\ge 1}\Lambda (n)n^{-s}$ giveApply an effective Perron formula on the line $\kappa =1+1/\log x$ and truncate atThe bound from part (c) controls the truncation error.

Use the classical Zero-free region of the Riemann zeta functiontogether with ${\zeta }^{\prime }/\zeta (s)\ll {\log }^2(|\Im s|+3)$ there. Contour shifting moves the Perron contour to $\Re s=1-c_1/\log T$. The only crossed singularity is the double pole at $s=1$, whose residue is $x\log x+Ax$ by part (b). On the new contour,and the logarithmic-derivative bounds contribute only powers of $\log x$, which can be absorbed by reducing the positive constant in the exponential. The horizontal integrals and Perron truncation error are $O(x\exp (-c_2\sqrt{\log x}))$ as well. Therefore, for some $c>0$,

Solved by gpt-5.6-sol high.