Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 124 / 1

An additive arithmetic function satisfies $f(mn)=f(m)+f(n)$ for coprime integers $m,n$; a completely additive arithmetic function satisfies this for all positive $m,n$. In either case $f(1)=0$. The fundamental theorem of arithmetic gives $f(n)={\sum }_{p^k\parallel n}f(p^k)$, where $p^k\parallel n$ means that the prime valuation of $n$ at $p$ is exactly $k$.

Under the discrete uniform distribution on $[N]$, the expectation of the indicator function of $p^k\parallel n$ isThe formula remains valid when $p^{k+1}>N$. Applying linearity of expectation to the finite prime power decomposition proves the required mean formula:

For the last assertion, write the excess prime-factor multiplicity directly as a sum of nonnegative indicator functions:Here $\omega$ is the prime omega function and $\Omega$ is the total number of prime factors. The expectation is bounded independently of $N$, sinceThus Markov inequality gives vanishing probability on every diverging scale:

The Möbius function has $\mu (1)=1$, $\mu (n)=0$ when $n$ is divisible by a prime square, and $\mu (n)=(-1{)}^r$ when $n$ is a product of $r$ distinct primes. The Riemann zeta function is initially the Dirichlet series $\zeta (s)={\sum }_{n\ge 1}{n}^{-s}$ for $\sigma =\Re s>1$. Both this Dirichlet series and $\sum \mu (n)n^{-s}$ have absolute convergence, because $|\mu (n)|\le 1$ and $\sum n^{-\sigma }<\infty$.

For a finite set of primes $p\le y$, multiplication of absolutely convergent geometric series and unique prime factorization giveEvery omitted integer has a prime factor exceeding $y$, so it exceeds $y$. The difference from $\zeta (s)$ has absolute value at most ${\sum }_{n>y}{n}^{-\sigma }\to 0$. This proves the Euler product, with the product interpreted as the limit of its finite products:

For the reciprocal identity, absolute convergence justifies multiplying and regrouping the two Dirichlet series by their product index. Their Dirichlet convolution coefficient at $m$ isThe coefficient calculation uses only the squarefree integers dividing $m$. Consequently the reciprocal identity is an identity of absolutely convergent series:

Let $M(x)={\sum }_{n\le x}\mu (n)$ be the Mertens function. Partial summation givesFor any $\Re s>0.9$, choose a sufficiently small $ϵ>0$ with $0.9+ϵ<\Re s$. The assumed bound makes the boundary term tend to zero and makes the integral absolutely convergent. Moreover, on each compact set of this half-plane, one can choose the same $ϵ$, so the integral has locally uniform convergence. Henceis a holomorphic function on $\Re s>0.9$, agreeing with the reciprocal Dirichlet series from part (b) on $\Re s>1$.

The identity theorem applied to the holomorphic functions $(s-1)\zeta (s)F(s)$ and $s-1$ extends their equality throughout $\Re s>0.9$; the assumed simple pole makes the first function holomorphic at $1$. Thus $\zeta (s)F(s)=1$ away from $1$, proving a zero-free half-plane:

To obtain the strongest symmetric conclusion, use the following basic facts explicitly: the functional equation of the Riemann zeta function gives the entire function $\xi (s)=\frac{1}{2}s(s-1){\pi }^{-s/2}\Gamma (s/2)\zeta (s)$ with $\xi (s)=\xi (1-s)$, and the trivial zeros of the Riemann zeta function are the negative even integers. Every Nontrivial zero of the Riemann zeta function is therefore accompanied by its reflection $1-s$. A Nontrivial zero of the Riemann zeta function with real part below $0.1$ would reflect to one above $0.9$, which is impossible. Consequently all nontrivial zeros lie in the closed stripThe $ϵ$ in the hypothesis prevents this argument from excluding either boundary line; the trivial zeros of the Riemann zeta function remain present.