∀ a, n ∈ N : ∣a−n∣=1 pour a, n ≥3 a^m ≡1modn (∗) posons : m=n−1 (∗′) subtituons cette valeur dans (∗). on a: a^(n−1) ≡1modn. Mais n n′est pas forcement premier. Test de primalite ∀ n ∈ N, n ≥3. (n−2)^(n−1) ≡1modn ⇒ n est premier.


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Question Number 61096 by Arthur El-bomart last updated on 29/May/19

$\forall a, n \in N : ∣ a - n ∣= 1 p o u r a, n ⩾ 3$ $a^{m} \equiv 1 m o d n ()$ $p o s o n s : m = n - 1 (^{'})$ $s u b t i t u o n s c e t t e v a l e u r d a n s (*) .$ $o n a : a^{n - 1} \equiv 1 m o d n . M a i s n n^{'} e s t p a s f o r c e m e n t p r e m i e r .$ $T e s t d e p r i m a l i t e$ $\forall n \in N, n ⩾ 3 .$ ${(n - 2)}^{n - 1} \equiv 1 m o d n \Rightarrow n e s t p r e m i e r .$
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