Lemma 1.65. Let p be an odd prime and e ≥ 2. For all a ∈ ℤ, we have
Proof: For e = 2, the statement is trivial. So, let e > 2. Inductively, we have 
≡ 1 + ape−2 mod pe−1. Hence, there is a number b ∈ ℤ such that 
+ ape−2 + bpe−1. Using 
for 1 ≤ k < p, we obtain
Now, because p > 2, and thus mod pe follows.
□
Lemma 1.66. Let e ≥ 3. Then for all a
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