{c}
{wiki=Riemann_zeta_function#Zeros}

A nontrivial zero of the <Riemann zeta function> lies in the <critical strip> $0<\Re s<1$. The <Functional equation of the Riemann zeta function> reflects such zeros across the <critical line> $\Re s=1/2$.

Nontrivial zero of the Riemann zeta function

Put
$$
A(x)=\sum_{n\leq x}\Lambda(n)n^{iu}
=\frac{x^{1+iu}}{1+iu}+O\left(x^{1/2}(\log x)^2\right).
$$
For $\Re s>1$, <partial summation> gives
$$
-\frac{\zeta'(s-iu)}{\zeta(s-iu)}
=s\int_1^\infty A(x)x^{-s-1}\,dx
=\frac{s}{(1+iu)(s-1-iu)}
+s\int_1^\infty O\left(x^{1/2}(\log x)^2\right)x^{-s-1}\,dx.
$$
The 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 $-\zeta'(s-iu)/\zeta(s-iu)$ 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>.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 150 / 2 / d / Solution

The symmetric form of the <Functional equation of the Riemann zeta function> is
$$
\xi(s)=\xi(1-s),
\qquad
\xi(s)=\frac12s(s-1)\pi^{-s/2}\Gamma(s/2)\zeta(s).
$$
The <complex conjugate>[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.

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 150 / 2 / b / Solution

Write $M(x)=\sum_{n\leq x}\mu(n)$ for the <Mertens function>. Suppose, to the contrary, that for some $\varepsilon>0$ the quotient $|M(x)|/x^{1/2-\varepsilon}$ were bounded. <Partial summation> would then make
$$
\sum_{n=1}^{\infty}\frac{\mu(n)}{n^s}
=s\int_1^\infty M(x)x^{-s-1}\,dx
$$
converge and define a <holomorphic function> throughout $\Re s>1/2-\varepsilon$. 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 of the Riemann zeta function>[nontrivial zero]. The <Functional equation of the Riemann zeta function> reflects one of that zero and its partner into $\Re s\geq1/2$, where $1/\zeta$ must have a pole, a contradiction. Thus $|M(x)|/x^{1/2-\varepsilon}$ is unbounded, which gives an $x\geq1$ exceeding any prescribed constant $C$.

Solved by gpt-5.6-sol high.

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 150 / 2 / c / Solution

{c}
{wiki=Riemann_zeta_function#Riemann's_functional_equation}

The completed zeta function
$$
\xi(s)=\frac12s(s-1)\pi^{-s/2}\Gamma(s/2)\zeta(s)
$$
is an <entire function> satisfying $\xi(s)=\xi(1-s)$. Thus every nontrivial zero $\rho$ is accompanied by $1-\rho$, $\overline\rho$, and $1-\overline\rho$.

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 Incoming links: Functional equation of the Riemann zeta function