Congruence Modulo Power of p as Linear Combination of Congruences Modulo p

Theorem
Let $p$ be a prime number.

Let $S = \set {a_1, a_2, \ldots, a_p}$ be a complete residue system modulo $p$.

Then for all integers $n \in \Z$ and non-negative integer $s \in \Z_{\ge 0}$, there exists a congruence of the form:


 * $n \equiv \ds \sum_{j \mathop = 0}^s b_j p^j \pmod {p^{s + 1} }$

where $b_j \in S$.

Proof
Proof by induction on $s$:

Basis for the Induction
For $s = 0$, we apply the definition of a complete residue system modulo $p$:
 * $\forall n \in \Z: \exists a_i \in S: n \equiv a_i \pmod p$

This is our base case.

Induction Hypothesis
This is our induction hypothesis:
 * For some $k \in \Z_{\ge 0}$, for all integers $n \in \Z$, there exists a congruence of the form:


 * $n \equiv \ds \sum_{j \mathop = 0}^k b_j p^j \pmod {p^{k + 1} }$

It is to be demonstrated that:
 * For all integers $n \in \Z$, there exists a congruence of the form:


 * $n \equiv \ds \sum_{j \mathop = 0}^{k + 1} b_j p^j \pmod {p^{k + 2} }$

Induction Step
This is our induction step:

From $n \equiv \ds \sum_{j \mathop = 0}^k b_j p^j \pmod {p^{k + 1} }$ we have:
 * $\exists q \in \Z: n = \ds \sum_{j \mathop = 0}^k b_j p^j + q p^{k + 1}$

From the definition of a complete residue system modulo $p$:
 * $\exists a_i \in S: q \equiv a_i \pmod p$

This gives:
 * $\exists r \in \Z: q = a_i + r p$

Substituting this to the original equation we have:

This shows that $n$ can be expressed as the form:
 * $n \equiv \ds \sum_{j \mathop = 0}^{k + 1} b_j p^j \pmod {p^{k + 2} }$

By the Principle of Mathematical Induction, the theorem is true for any $s$.

Note that in the proof above, we did not use the fact that $p$ is a prime number.