The Mertens function $M(x)$ is defined as $\Big|\sum_{n\leq x} \mu(n)\Big|$, where $\mu$ denotes the Mobius function.
If $\chi$ is a primitive character modulo $q$, the Polya-Vinogradov inequality states that
$$|\sum_{n\leq q} \chi(n) | \ll q^{1/2}\log q.$$
Note that the Mobius function can be viewed as a primitive character except that
1.) It is aperiodic,
2.) It is not completely multiplicative, thus is not a primitive character in the usual sense.
However,
1.)The lack of a period for $\mu$ can be interpreted by saying $\mu$ has an ``infinite'' period.
2.) Also, what makes $\mu$ not to be completely multiplicative is the fact that for squarefree $n$, $\mu^{2}(n)=1\neq \mu(n^2)=0$. However, the good news is that the $\mu$ vanishes at the squares, so the squares $\leq x$ have no contribution to $M(x)$.
These two observations allow us to apply the Polya-Vinogradov inequality to $M(x)$ with the period for $\mu$ being $q\leq x$.
We thus arrive at
$$M(x) \ll x^{1/2}\log x $$
which proves the Riemann Hypothesis. Is there anything wrong here ?