Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 150 (Analytic Number Theory)

WriteThe sifting function isFor each integer $a$, Möbius inversion in its divisor-indicator form givesSumming over the finite set $A$ and interchanging the finite sums yieldsThis is the inclusion-exclusion formula encoded by the Möbius function.

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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.

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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 $\ll x/(\log x{)}^3$.

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The Truncated Perron formula says that if $F(s)={\sum }_{n\ge 1}{a}_n{n}^{-s}$ converges absolutely for $\Re s>{\sigma }_a$, then for $c>{\sigma }_a$, $T\ge 2$, and $x$ not an integer,Take $a_n=\Lambda (n)$ and $c=1+1/\log x$. The logarithmic derivative identity gives $F(s)=-{\zeta }^{\prime }(s)/\zeta (s)$. Since $x-1/2$ is an integer, $|x-n|\ge 1/2$ for every integer $n$. In the range $x/2<n<2x$,and hence the contribution there isThe ranges $n\le x/2$ and $n\ge 2x$ are bounded by the same quantity using absolute convergence and $-{\zeta }^{\prime }(c)/\zeta (c)\ll \log x$. Therefore

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Let $L=\log x$ and choosewith fixed positive $a,b$ chosen so that the rectangle up to height $T$ lies inside the Zero-free region of the Riemann zeta function. Move the Perron contour from $1+1/L$ to ${\sigma }_0$. The only singularity crossed is the simple pole at $s=1$ of $-{\zeta }^{\prime }(s)/\zeta (s)$, whose residue contributes $x$.

The standard bound ${\zeta }^{\prime }(s)/\zeta (s)\ll (\log T{)}^2$ in this zero-free rectangle givesThe two horizontal sides are $\ll x(\log T{)}^2/T$, and the truncation error from part a is $\ll x{L}^2/T$. Polynomial factors in $L$ can be absorbed by slightly reducing the exponential constant. Thus some $c>0$ satisfiesThis is the Prime number theorem with classical zero-free-region error.

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For $\Re s>1$, absolute convergence permits a shift of the complex variable:

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PutFor $\Re s>1$, partial summation givesThe last integral is holomorphic for $\Re s>1/2$. Thus the logarithmic derivative on the left continues meromorphically to that half-plane with no pole except $s=1+iu$.

A zero $\rho$ of $\zeta$ with $\Re \rho >1/2$ would make $-{\zeta }^{\prime }(s-iu)/\zeta (s-iu)$ singular at $s=\rho +iu$, a contradiction unless $\rho =1$, which is a pole rather than a zero. Therefore no nontrivial zero lies to the right of the critical line. The Functional equation of the Riemann zeta function reflects zeros across that line, so none lies to its left either. Every nontrivial zero lies on the critical line, proving the Riemann hypothesis. This is the Twisted Von Mangoldt estimate implying the Riemann hypothesis.

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For $\sigma =\Re s>1$,On every compact subset of $\Re s>1$, the terms are bounded by a convergent series $\sum n^{-1-\delta }$. The Weierstrass M-test gives locally uniform convergence, and the theorem on locally uniform convergence of holomorphic functions shows that the limit is an analytic function. Hence $D_f$ is analytic throughout $\Re s>1$.

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For arithmetic functions $f,g$ bounded in modulus by one, their pretentious distance up to $x$ is defined by

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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

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For $\sigma >1$, take logarithms of Euler products. With $z=f(p)p^{-it}$, the inequalityat every prime power gives the three-four-one inequality for Euler products

Suppose $D_f(1+it)=0$ with multiplicity $m\ge 1$. The assumed analytic continuation giveswhile $D_{f^2}(\sigma +2it)$ remains bounded and $\zeta (\sigma {)}^3\asymp (\sigma -1{)}^{-3}$ as $\sigma \downarrow 1$. The left side of the inequality would then becontradicting its lower bound one. Hence $D_f$ has no zero on $\Re s=1$. Absolute convergence of its Euler product already excludes zeros for $\Re s>1$, so $D_f(s)\ne 0$ throughout $\Re s\ge 1$.

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