Solving Congruences mod p^e

Suppose that p is prime and e>1. If xis a solution of a congruence f(x)=0(modpe), then x satisfies f(x)≡0(modp), so one way of finding all solutions of f(x)≡0(modpe) is first to solve the simpler congruence f(x)=0(modp),and then to see which solutions of this are also solutions of the more restrictive congruence f(x)≡0(modpe). In many cases an effective strategy is to increase the exponent of p one step at a time, solving f(x)≡0(modp), then f(x)≡0(modp2), and so on until we reach the modulus pe. 
 
Example:

We could have solved the congruence in Example more directly by writing it as 2x≡3+5e(mod5e); since 3+5e is even, and 2 is coprime to 5^e,we get the general solution c=(3+5e)/2(mod5). Instead, we used the longer iterative method to give a simple illustration of how this method works.

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