Problem 1582

We will use the well-known fact that $\nu(n) = \sum_{k=1}^{\infty} \lfloor \frac{n}{2^k} \rfloor$. If the base-2 form of the number $n$ is $n = \overline{d_l d_{l-1} \dots d_1 d_0} = \sum_{i=0}^{\infty} d_i \cdot 2^i$, where the digits $d_0, \dots, d_l$ are all $0$ or $1$, then
$$\begin{aligned} \nu(n) &= \sum_{k=1}^{\infty} \lfloor \frac{n}{2^k} \rfloor = \sum_{k=1}^{l} \lfloor \frac{\overline{d_l d_{l-1} \dots d_1 d_0}}{2^k} \rfloor = \sum_{k=1}^{l} \overline{d_l d_{l-1} \dots d_k} \\ &= \sum_{k=1}^{l} \left( \sum_{i=k}^{l} d_i \cdot 2^{i-k} \right) = \sum_{i=1}^{l} d_i \left( \sum_{k=1}^{i} 2^{i-k} \right) = \sum_{i=1}^{l} d_i \cdot (2^i - 1). \end{aligned}$$
Now we will show that there exists an integer $r$ which is relatively prime to $m$, and an infinite sequence $i_1 < i_2 < i_3 < \dots$ of positive integers such that
$$2^{i_1} - 1 \equiv 2^{i_2} - 1 \equiv 2^{i_3} - 1 \equiv \dots \equiv r \mod m.$$
Let $m = 2^t u$ where $u$ is odd, and consider an arbitrary positive integer $i \ge t$ for which $\varphi(u)$ divides $i-1$. By the Euler-Fermat theorem,
$$u \mid 2^{\varphi(u)} - 1 \mid 2^{i-1} - 1,$$
$$2^i - 1 = 2(2^{i-1} - 1) + 1 \equiv 1 \quad \text{mod } u,$$
and, due to $i \ge t$,
$$2^i - 1 \equiv -1 \quad \text{mod } 2^t.$$
The relations (1) and (2) determine the residue class of $2^i - 1$ modulo $m$, and it must be relatively prime to $m$.
Since $m$ and $r$ are relatively prime, there is a positive integer $u$ such that $a \equiv u r \mod m$. For $n = 2^{i_1} + \cdots + 2^{i_u}$ we achieve
$$\nu(n) = \nu(2^{i_1} + \cdots + 2^{i_u}) = (2^{i_1} - 1) + \cdots + (2^{i_u} - 1) \equiv u \cdot r \equiv a \quad \text{mod } m.$$