Given $P(x) \in \mathbb{Z}[x]$ a polynomial, $n$ a positive integer and $p$ a prime, there is a result that relates $|\text{Im } P \pmod{p^n}|$ with $|\text{Im } P \pmod{p^{n+1}}|$ perhaps in terms of some divisibility?
Obs. Clearly, $\text{Im } P \pmod{n} = \{ m \in \mathbb{Z}_n \ :\ \exists\ x \in \mathbb{Z}_n \mid P(x)\equiv m \pmod n\}$.
The motivation of this question is just looking at the polynomial $x^k$, $k \in \mathbb{Z}^*$ modulo $p^n$ and $p^{n+1}$ and noticing the divisibility $$\#(\text{Im } P \pmod{p^n})- l_n \mid \# (\text{Im } P \pmod{p^{n+1}})- l_{n+1}$$ where $$l_{n} = \# \{m \in ( \mathbb{Z}_{p^n}-(\mathbb{Z}/p^n\mathbb{Z})^{\times}) \ :\ \exists\ x \in \mathbb{Z}_{p^n} \mid P(x)\equiv m \pmod{p^{n}} \},\ $$ $$l_{n+1} = \# \{m \in ( \mathbb{Z}_{p^{n+1}}-(\mathbb{Z}/p^{n+1} \mathbb{Z})^{\times}) \ :\ \exists\ x \in \mathbb{Z}_{p^{n+1}} \mid P(x)\equiv m \pmod{p^{n+1} }\}$$ given when considering $x$ written with the primitive root modulo $p^{n+1}$ and the outputs when $x \in (\mathbb{Z}/p^n\mathbb{Z})^{\times}$ and $x \in (\mathbb{Z}/p^{n+1}\mathbb{Z})^{\times}$.
