# Does the Mertens function have an infinite number of zeros?

The Mertens function is defined as follows:

(1) $$\quad\mathcal{M}(N)=\sum_{n=1}^N\mu(n)$$

I have a very simple question for which I cannot seem to find a definitive answer on the web. I've also consulted several books on number theory and can't even seem to find a conjecture much less a proven result.

Question 1: Does the Mertens function $$\mathcal{M}(N)$$ evaluate to zero for an infinite number of integers $$N$$?

I also have a second related question.

Question 2: What is the largest known integer N such that $$\mathcal{M}(N)=0$$?

• See OEIS sequence A002321 and references there. According to a comment by Charles R Greathouse, all integers appear infinitely often in this sequence. – Robert Israel Jul 3 '17 at 21:54
• – Jack D'Aurizio Jul 3 '17 at 22:31
• The Mertens function was computed at every positive integer $x \leq 10^{16}$ here. In this interval there are $366567325$ (integer) zeros and the largest one is $9511908657769636$. – Chip Hurst Jan 4 '18 at 21:56

Assume $M(x)$ has constant sign for $x > A$. Then $$\frac{1}{s \zeta(s)} = \int_1^\infty M(x)x^{-s-1}dx =g(s)+ \int_A^\infty M(x)x^{-s-1}dx$$ where $g(s) = \int_1^A M(x) x^{-s-1}dx$ is entire.
Hence $$|\frac{1}{s \zeta(s)}-g(s)| \le\int_A^\infty |M(x)x^{-s-1}|dx = \pm \int_A^\infty M(x)x^{-\sigma-1}dx =|\frac{1}{\sigma \zeta(\sigma)}-g(\sigma)|$$ Proving $\frac{1}{s \zeta(s)}-g(s)$ is analytic on $\Re(s) > \sigma$ and has a singularity at $s= \sigma$, where $\sigma$ is the abscissa of convergence of the integral (this would make the RH very easy to prove or disprove !)
But we know $\frac{1}{s \zeta(s)}$ is analytic on $(0,\infty)$ and has a pole at $s \approx 1/2+i14.134725$
• Interesting proof. I have a further question: assuming RH and given the informations about the distributions of zeroes of $\zeta$ along the critical line provided by the Riemann-Von Mangoldt theorem, is it possible to estimate how often the Mertens function has to exhibit a sign change? – Jack D'Aurizio Jul 3 '17 at 23:17
• @JackD'Aurizio I would try using the explicit formulas $$\psi(x)-x-\log 2\pi = \sum_{|Im(\rho)| < T} \frac{x^\rho}{\rho}+ \mathcal{O}(x^{1/2} \log^2x/T),\qquad M(x)+2 = \sum_{|Im(\rho)| < T} \frac{x^\rho}{\rho \zeta'(\rho)}+ \mathcal{O}(x^{1/2} \log^2x/T)$$ (if the zeros are simple) – reuns Jul 3 '17 at 23:34
• @JackD'Aurizio Using the density of zeros, we can try finding a simple function $h(x)$ such that $(\psi(x)-x)h(x)$ is almost always non-negative, so that $\int_1^\infty (\psi(x)-x) h(x) x^{-s-1}dx$ has a singularity at $s =\sigma$ the abscissa of convergence of $\int_1^\infty (\psi(x)-x) x^{-s-1}dx$ – reuns Jul 3 '17 at 23:40