Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 27 / 4

Let $1(n)=1$ and $ϵ(n)=1_{n=1}$, the identity for Dirichlet convolution. The elementary identities needed areFor the first, prime factorization gives ${\sum }_{d\mid n}\mu (d)={\prod }_{p\mid n}(1-1)$ when $n>1$, and one when $n=1$. For the second, if $n=\prod p^{a_p}$, then ${\sum }_{d\mid n}\Lambda (d)={\sum }_p{a}_p\log p=\log n$, the Von Mangoldt divisor identity. Convolving the second identity with $\mu$ proves the third.

For $U,V\ge 1$, let the subscripts $\le U$, $>U$, $\le V$, $>V$ denote truncations. The Vaughan identity isIts pointwise form isTo prove it, use ${\mu }_{>U}\ast 1=ϵ-{\mu }_{\le U}\ast 1$ and $1\ast {\Lambda }_{>V}=\log -1\ast {\Lambda }_{\le V}$. This givesand rearrangement proves the formula, as in the Vaughan identity proof.

The Bombieri–Vinogradov theorem states that for every $A>0$ there is $B>0$ such thatHere $\psi (y;q,a)={\sum }_{\begin{array}{c}n\le y\\ n\equiv a(\mathrm{mod}q)\end{array}}\Lambda (n)$ is the Chebyshev function in an arithmetic progression, and $\varphi$ is the Euler totient function. In particular, partial summation givesafter choosing the logarithmic saving in the weighted theorem sufficiently large and absorbing the prime power terms. This is the offset logarithmic integral function.

For the proof strategy, Orthogonality of Dirichlet characters converts errors in arithmetic progressions into character sums of Dirichlet characters weighted by $\Lambda$. Reduction to primitive Dirichlet characters is followed by the character large sieve, which bounds their mean square by $O((N+Q^2)\sum |a_n{|}^2)$ with the usual $q/\varphi (q)$ weights. The Vaughan identity separates the weighted sums into short terms, Type I sums and Type II sums. Direct inner-sum estimates handle the Type I sums; the Cauchy-Schwarz inequality and the large sieve control the Type II sums. The Siegel–Walfisz theorem supplies arbitrarily strong logarithmic savings for small moduli, where an average estimate alone would not suffice. A dyadic decomposition, suitable $U,V$ and a sufficiently large $B$ absorb the divisor and logarithmic losses. This explains why the range is essentially $Q^2\le x$, with logarithmic room to spare.

Put $E_q={\max }_{(a,q)=1}|\pi (x;q,a)-\mathrm{Li}(x)/\varphi (q)|$, with $\mathrm{Li}$ the offset logarithmic integral function. The permitted standard sieve estimate can be taken as the Brun–Titchmarsh theorem:For $q\le x^{0.99}$, $\log (x/q)\ge 0.01\log x$. Also $\mathrm{Li}(x)\ll x/\log x$. Thus both terms defining $E_q$ are $O(x/(\varphi (q)\log x))$, uniformly over the permitted moduli. Apply the Cauchy-Schwarz inequality in the formThis is the weighted arithmetic-progression error bound. The prime omega function and Euler totient function weights can therefore be handled using an unweighted mean error and a separate positive sum.

Take sieve level $D=x^{1/3}$ and coefficient cutoff $L=\sqrt{D}=x^{1/6}$. Let $\mathcal{P}$ be the finite set of odd primes at most $L$. Give $n=p+2$ one unit of weight for each prime $p\le x$. For an odd squarefree integer $d$ on these primes, the sieve distribution isThe residue $-2$ is coprime to every such $d$, and $g(p)=1/(p-1)\in (0,1)$ for all permitted primes. Values of $g$ outside the permitted set can be chosen arbitrarily in $(0,1)$ at primes and extended multiplicatively on squarefree integers.

For the Selberg sieve normalizing sum, $k(p)=g(p)/(1-g(p))=1/(p-2)$ andHere is a direct proof of this lower bound. Restrict the Euler product to $3\le p\le y=L^{1/4}$. The Mertens first theorem gives ${\sum }_{3\le p\le y}\log p/(p-1)=\log y+O(1)$, which is at most $\frac{1}{2}\log L$ for large $L$. The Mertens second theorem givesThe truncated Euler-product lower bound, with $\theta =1/2$, now gives $J\ge W/2\gg \log L$.

Use the minimizing Selberg sieve weights from Question 3. Their absolute values are at most one, so the Selberg least-common-multiple weights satisfy $|{\lambda }_d|\le 3^{\omega (d)}$. Parts (a) and (b) control the remainder:The first factor uses the Bombieri–Vinogradov theorem; $D=x^{1/3}$ is below $x^{1/2}/{\log }^{B}x$ for all sufficiently large $x$. The second uses the permitted ${\sum }_{q\le D}{9}^{\omega (q)}/\varphi (q)\ll {\log }^{9}D$. The PDF has the exponent nine, which the TeX transcribes incorrectly. Choose $A=14$ to obtain $O(x/{\log }^{3}x)$.

Every twin prime pair with $p+2>L$ survives the finite sieve, while the pairs with $p+2\le L$ number at most $L$. The Selberg upper-bound sieve thus givesTherefore the twin-prime count is $O(x/{\log }^{2}x)$. The finite cutoff on the forbidden primes is essential: sieving by all odd primes would also discard the large prime values $p+2$ that we are trying to count.