The Lumenary · First-year graduate

Analytic number theory

Name: ______________________

Date: ______________________

You will learn to

Manipulate arithmetic functions and Dirichlet series, including Dirichlet convolution, Möbius inversion and the Euler product for $\zeta(s)$
Prove Chebyshev's bounds $\psi(x) \asymp x$ and Euler's theorem that $\sum 1/p$ diverges
Continue $\zeta(s)$ to $\operatorname{Re} s > 0$ , explain the functional equation, and describe how the zeros of $\zeta$ control the primes through the explicit formula
Prove that $\zeta$ has no zeros on the line $\operatorname{Re} s = 1$ , outline the proofs of the prime number theorem and Dirichlet's theorem, and separate proved results from conjectures such as the Riemann hypothesis

1.
Compute $\mu(30) + \mu(36) + \mu(7)$ , where $\mu$ is the Möbius function.
2.
Which Dirichlet series equals $1/\zeta(s)$ for $\operatorname{Re} s > 1$ ?
- A. $\sum_n \mu(n)\, n^{-s}$
- B. $\sum_n (-1)^n n^{-s}$
- C. $\sum_n \lambda(n)\, n^{-s}$ with $\lambda(n) = (-1)^{\Omega(n)}$ (Liouville)
- D. $\sum_n \Lambda(n)\, n^{-s}$
3.
The prime number theorem says $\pi(x) \sim x/\log x$ . Which statement is a correct reading of it?
- A. $\pi(x) \big/ (x/\log x) \to 1$ as $x \to \infty$ .
- B. $\pi(x) - x/\log x \to 0$ as $x \to \infty$ .
- C. $\pi(x) = \lfloor x/\log x \rfloor$ for all large $x$ .
- D. $\pi(x) - \operatorname{li}(x) \to 0$ .
4.
In the proof that $\zeta(1 + it) \ne 0$ (Theorem 7), why does $\zeta(\sigma + 2it)$ stay bounded as $\sigma \to 1^+$ (for fixed $t \ne 0$ )?
- A. $\zeta$ is holomorphic at $1 + 2it$ , because its only pole is at $s = 1$ and $1 + 2it \ne 1$ .
- B.Because $|\zeta(\sigma + 2it)| \le \zeta(\sigma)$ .
- C.Because $\zeta$ has no zeros on $\operatorname{Re} s = 1$ .
- D.Because of the functional equation relating $s$ and $1 - s$ .
5.
Gauss's identity $\sum_{d \mid n}\varphi(d) = n$ says $\varphi * \mathbf{1} = \operatorname{id}$ . Möbius inversion gives $\varphi(n) = \sum_{d \mid n}\mu(d)\,\frac{n}{d} = n\prod_{p \mid n}\left(1 - \frac1p\right)$ . Compute $\varphi(60)$ .
6.
Every non-trivial zero $\rho$ of $\zeta$ with $0 < \operatorname{Im}\rho \le 3 \cdot 10^{12}$ has been verified to lie on the critical line. What follows?
- A.This is strong evidence for the Riemann hypothesis, but not a proof: it says nothing about the infinitely many zeros higher up.
- B.The Riemann hypothesis is proved for all practical purposes.
- C.By the functional equation, the remaining zeros must also be on the line.
- D.The prime number theorem would fail if the verification had found a zero off the line.
7.
Suppose (hypothetically) that $\zeta$ had a zero $\rho = 0.9 + i\gamma$ . Using the explicit formula, what is the best possible error term one could then hope for in $\psi(x) = x + E(x)$ ?
- A. $E(x)$ would be as large as $x^{0.9 - \varepsilon}$ for arbitrarily large $x$ , so no bound $O(x^{\Theta})$ with $\Theta < 0.9$ could hold.
- B. $E(x) = O(\sqrt x\log^2 x)$ still holds.
- C. $\psi(x) \sim x$ would fail.
- D.No effect: a single zero contributes a bounded term.
8.
Which Lean proposition correctly states the Riemann hypothesis (for Mathlib's riemannZeta : ℂ → ℂ, which assigns a junk value at the pole $s = 1$ )?
- A.∀ s : ℂ, riemannZeta s = 0 → (¬ ∃ n : ℕ, s = -2 * (n + 1)) → s ≠ 1 → s.re = 1 / 2
- B.∀ s : ℂ, riemannZeta s = 0 → s.re = 1 / 2
- C.∀ s : ℂ, riemannZeta s = 0 → (¬ ∃ n : ℕ, s = -2 * (n + 1)) → 0 < s.re ∧ s.re < 1
- D.∃ s : ℂ, riemannZeta s = 0 ∧ s.re = 1 / 2
9.
Write mobius(n) for $n \ge 1$ : return $0$ if $n$ has a repeated prime factor, and $(-1)^k$ if $n$ is a product of $k$ distinct primes. Then check numerically that $\sum_{d \mid n}\mu(d) = 0$ for $1 < n \le 200$ (Proposition 1).
Computational — Practice

Answer key — Analytic number theory

1.
$\mu(30) = (-1)^3 = -1$ , $\mu(36) = 0$ (as $4 \mid 36$ ), $\mu(7) = -1$ . The sum is $-2$ .
2.
Since $\mu * \mathbf{1} = \varepsilon$ (Proposition 1) and Dirichlet series multiply (Proposition 2), $\big(\sum \mu(n)n^{-s}\big)\zeta(s) = 1$ . Alternatively, invert the Euler product: $1/\zeta(s) = \prod_p (1 - p^{-s})$ , and expanding gives $\sum_n \mu(n)n^{-s}$ .
3.
$f \sim g$ means $f(x)/g(x) \to 1$ . The prime number theorem says nothing about the difference tending to zero; in fact $\pi(x) - x/\log x \sim x/\log^2 x \to \infty$ , because $\pi(x) = \frac{x}{\log x} + \frac{x}{\log^2 x} + O\big(\frac{x}{\log^3 x}\big)$ .
4.
By Theorem 5, $\zeta$ is meromorphic on $\operatorname{Re} s > 0$ with a single pole at $s = 1$ . For $t \ne 0$ the point $1 + 2it$ is not that pole, so $\zeta$ is holomorphic, in particular continuous, there, and $\zeta(\sigma + 2it) \to \zeta(1 + 2it)$ . (This is why the argument needs $t \neq 0$ ; at $t = 0$ , $\zeta$ has a pole, not a zero.)
5.
$\varphi(60) = 60 \cdot \frac12 \cdot \frac23 \cdot \frac45 = 16$ . Via the sum: $\sum_{d \mid 60}\mu(d)\frac{60}{d} = 60 - 30 - 20 - 12 + 10 + 6 + 4 - 2 = 16$ , the terms coming from $d = 1, 2, 3, 5, 6, 10, 15, 30$ .
6.
There are infinitely many non-trivial zeros (about $\frac{T}{2\pi}\log\frac{T}{2\pi}$ up to height $T$ ), so checking any finite number of them proves nothing about the rest. The verification (Platt and Trudgian, 2021) is evidence; the Riemann hypothesis remains open. The prime number theorem, by contrast, is proved, because it requires only $\zeta(1 + it) \ne 0$ .
7.
$|x^\rho/\rho| = x^{\beta}/|\rho|$ , so a zero with $\beta = 0.9$ contributes oscillations of size $x^{0.9}$ . One can show (Landau, Ingham) that if $\Theta = \sup\operatorname{Re}\rho$ then $\psi(x) - x = O(x^{\Theta}\log^2 x)$ and $\psi(x) - x \ne o(x^{\Theta - \varepsilon})$ for every $\varepsilon > 0$ . So the error would be at least $x^{0.9 - \varepsilon}$ infinitely often — still $o(x)$ , so the prime number theorem would survive, but the Riemann hypothesis ( $\Theta = \frac12$ ) would be false.
8.
The Riemann hypothesis says every non-trivial zero has real part $\frac12$ . In Lean, 'non-trivial' means excluding $s = -2(n + 1)$ , and the condition $s \ne 1$ guards against Mathlib's junk value at the pole. The statement without the exclusion is false; the strip statement is a theorem; the existential statement is easy. Lean can state the conjecture exactly — but no one can prove it.
9.
```
def mobius(n):
    k, m, p = 0, n, 2
    while p * p <= m:
        if m % p == 0:
            m //= p
            if m % p == 0:
                return 0          # p^2 divides n
            k += 1
        p += 1
    if m > 1:
        k += 1                   # one prime factor left over
    return (-1) ** k
```
The starter computed Liouville's function $\lambda(n) = (-1)^{\Omega(n)}$ , which agrees with $\mu$ on squarefree numbers but not elsewhere ( $\lambda(12) = -1$ , $\mu(12) = 0$ ); its Dirichlet series is $\zeta(2s)/\zeta(s)$ , not $1/\zeta(s)$ . The divisor-sum check confirms $\mu * \mathbf{1} = \varepsilon$ on a range — evidence for an identity that Proposition 1 proves for all $n$ . The partial sums $M(x) = \sum_{n \le x}\mu(n)$ (here $M(100) = 1$ ) are tied to the zeros: $M(x) = O(x^{1/2 + \varepsilon})$ is equivalent to the Riemann hypothesis.