Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 150 (Analytic Number Theory)

The Von Mangoldt divisor identity givesBy the Stirling formula, the left side is $\log (\lfloor x\rfloor !)=x\log x-x+O(\log x)$, while the assumed Chebyshev estimate gives $\psi (x)=O(x)$. HenceThe contribution of proper prime powers is bounded uniformly:Removing it leaves

PutThe Abel summation formula with weight $1/\log t$ givesSubstituting $A(t)=\log t+E(t)$ yieldsfor a constant $c$. Both final terms are $O(1/\log x)$, soThis is the Mertens second theorem.

The same divisor-identity argument as in part (a) givesOn the other hand, the Abel summation formula givesUnder the proposed asymptotic, the right side isDividing first by $\log x$ gives $a=1$. Subtracting $\log x$, dividing by $\log \log x$, and taking the limit then gives $b=0$. Therefore

Taking the natural logarithm of the finite Euler product and using the Taylor seriesgivesThe double series converges absolutely, and the Mertens second theorem therefore makes the right sideforExponentiating provesThis is the Mertens third theorem.

The prime-factor formula for the Euler totient function isLetThe factors with $p\le y$ contribute at mostby the Mertens third theorem. For $p>y$, the number of distinct prime divisors of $n$ is at most $\log n/\log y$, and henceThus the large-prime product is $1+o(1)$, andTaking reciprocals gives, uniformly as $n\to \infty$,

Let $\mathcal{A}=(a_n)$ be a finite nonnegative sequence, let $\mathcal{P}$ be a set of prime numbers, and defineThe sifting function isFor a finite set $A$ of integers, take $a_n$ to be the number of occurrences of $n$ in $A$; then $S(A,\mathcal{P};z)$ counts members divisible by no $p\in \mathcal{P}$ below $z$. For $d\mid P(z)$, write

Suppose the sieve distribution has the formwhere $g$ is a multiplicative arithmetic function on squarefree integers and $0\le g(p)<1$. DefineThen the Selberg upper-bound sieve statesfor $1<D\le z$, with immaterial endpoint changes under other level conventions.

To construct the weights, putand setThen ${\lambda }_1=1$. If $\gcd (n,P(z))=1$, the divisor sum ${\sum }_{d\mid n}{\lambda }_d$ equals one, and henceSumming against $a_n$ and expanding givesThe Selberg diagonalization of the positive quadratic form givesFinally, grouping the error by $m=[d,e]$ gives at most $3^{\omega (m)}$ pairs $(d,e)$ for each squarefree $m$; using $|{\lambda }_d|\le 1$ yields the stated remainder.

Letand let $W$ be the product of the finitely many primes at most $r$ or dividing some nonzero difference $h_i-h_j$. Sieve with the primes $p\nmid W$. For such a prime, the congruence $F(n)\equiv 0(\mathrm{mod}p)$ has exactly $r$ distinct roots. By the Chinese remainder theorem, the root-counting function $\rho (d)$ is multiplicative on squarefree $d$ coprime to $W$, andThus the polynomial root density in a sieve applies with

Take $z=x^{1/4}$ and $D=z$. For squarefree $d$ coprime to $W$,The supplied mean-value estimate, partial summation, and removal of square factors using $r^{\omega (n)}{\ll }_{r,ϵ}{n}^{ϵ}$ giveThe error in the Selberg upper-bound sieve isConsequently

If every $n+h_i$ is prime and all of them exceed $z$, then $F(n)$ has no prime divisor $p<z$ with $p\nmid W$, apart from a fixed finite set of divisibility cases absorbed into the implied constant. The cases with some $n+h_i\le z$ contribute $O_r(z)$. Therefore

For $\sigma >1$, summation over the intervals $[n,n+1)$ givesSince $\lfloor u\rfloor =u-\{u\}$,The integral converges absolutely and defines a holomorphic function for $\sigma >0$. This formula therefore gives the Meromorphic continuation of the Riemann zeta function to the right half-plane, with only a simple pole at $s=1$ and residue one.

For $\sigma >1$, both relevant Dirichlet series converge absolutely, so their product may be rearranged:The inner sum is one when $n=1$ and zero otherwise by Möbius inversion. HenceThis is also the reciprocal of the absolutely convergent Euler product.

For $\sigma >1$, the three-four-one zero-free-region argument starts fromApplying this termwise togives

Suppose $\rho =\beta +i\gamma$ is a zero with $|\gamma |\ge 4$. In the supplied Local partial-fraction expansion of the Riemann zeta logarithmic derivative, every nearby zero contributes a nonnegative real part at $\sigma +i\gamma$ when $\sigma >1$. Keeping the term from $\rho$ givesAt height zero the pole at one givesand the same local expansion at height $2\gamma$ givesSubstitution yieldsSet $\sigma =1+\eta /\log |\gamma |$, first choosing a sufficiently small absolute $\eta >0$. If $\beta >1-c/\log |\gamma |$, the left side is at least $4\log |\gamma |/(\eta +c)$. Choosing $c>0$ sufficiently small contradicts the last inequality. Therefore

The truncated Perron formula says that, for $c>1$ and $T\ge 2$,where an endpoint has half weight and the error is controlled by

Apply this with $a_n=\mu (n)$, $c=1+1/\log x$, andPart (b) makes the integrand $x^s/(s\zeta (s))$. Move the contour tousing a contour that stays inside the Zero-free region of the Riemann zeta function near small $|t|$. The estimates supplied in the question give $1/\zeta (s)\ll \log (|t|+3)$ on the new contour. Its vertical segment is thereforeafter decreasing $c_3$. The horizontal segments and the Perron truncation error areNo residue is crossed because $1/\zeta (s)$ has a zero, rather than a pole, at $s=1$. Thus the Mertens function satisfiesfor some absolute $c>0$.

Let $\hat{G}$ be the group of all Dirichlet characters modulo $q$, where $G=(\mathbb{Z}/q\mathbb{Z}{)}^{\times }$, and considerFor $p\nmid q$, let $f_p$ be the order of $p$ in $G$. The values $\chi (p)$ run through the $f_p$th roots of unity, each $|G|/f_p$ times, so the local factor isFor $p\mid q$ the local factor is one. Thus the Dirichlet series for $F$ has nonnegative coefficients.

The principal-character factor has a simple pole at $s=1$, while every nonprincipal Dirichlet L-function is entire. If some nonprincipal $L(1,\chi )$ vanished, its zero would cancel that pole and make $F$ entire. The Landau theorem for a Dirichlet series with nonnegative coefficients would then force the Dirichlet series of $F$ to converge for every real $s$. This is impossible: its coefficient at $m^{|G|}$ is at least one for every $m$ coprime to $q$, as is clear from the local factors. Hence

For $\sigma >1$, Orthogonality of Dirichlet characters givesThe principal-character term isPart (a) makes every nonprincipal logarithmic derivative bounded as $\sigma \to 1^{+}$, soIf the nondecreasing Chebyshev function in an arithmetic progressionwere bounded, the Abel summation formula would keep the displayed Dirichlet series bounded near $\sigma =1$. Therefore

There is an absolute $c>0$ such that the product of the Dirichlet L-functions modulo $q$ has no zero inexcept possibly one zero. If it exists, this exceptional zero $\beta$ is real and simple, belongs to a real nonprincipal character ${\chi }_1$, and lies very close to one. Among the primitive characters whose conductors divide $q$, at most one can have such a zero. This is the Classical zero-free region for Dirichlet L-functions.

Let ${\chi }_1$ be the real character associated with the exceptional zero $\beta$. The prime number theorem in an arithmetic progression with an exceptional zero gives, uniformly for $x\ge 2$ and $(a,q)=1$,If no exceptional zero exists, the middle term is omitted. The constant $c>0$ is absolute.

Assume $\beta$ exists and choose a reduced residue class $a$ with ${\chi }_1(a)=-1$; such a class exists because ${\chi }_1$ is nonprincipal. Fix a sufficiently large constant $A=A(ϵ)$ and takeThen $x\ge q^2$ for large $q$. Multiplying the formula from part (d) by $\phi (q)/x$ givesChoose $A$ so large that the error is at most $ϵ/4$ for all sufficiently large $q$. The assumed upper bound then impliesSince $\beta <1$, this givesTaking logarithms,Therefore, with a positive constant depending only on $ϵ$,