Let $\sigma_0(n)$ be the divisor counting function: $$\sigma_0(n) = \sum_{d \vert n} 1.$$ I ran some numerical experiments that showed when $p$ is prime, the function $\sigma_0(n)$ is equidistributed mod $p$. That is, for any residue class $a \mod p$, $$\lim_{X \to \infty} \dfrac{ \vert \{ n<X: \sigma_0(n) \equiv a \mod p \} \vert }{X} = \dfrac{1}{p}.$$

Is this fact already known? If so, could anyone sketch a proof / provide a reference for a proof?

Let $\sigma_0(n)$ be the divisor counting function: $$\sigma_0(n) = \sum_{d \vert n} 1.$$ I ran some numerical experiments that showed when $p$ is prime, the function $\sigma_0(n)$ is equidistributed mod $p$. That is, for any residue class $a \mod p$, $$\lim_{X \to \infty} \dfrac{ \vert \{ n<X: \sigma_0(n) \equiv a \mod p \} \vert }{X} = \dfrac{1}{p}.$$

Is this fact correct? If so, could anyone sketch a proof / provide a reference for a proof?