22 fév. 2005 (i) a ? b (mod n) implies ak ? bk (mod n) for all k ? 0. Solution. We use induction. ... holds because both sides are congruent to 0.
a ? 0 (mod n) ?? a = nk avec k entier. ?? a est un multiple de n. Quelques propriétés de la congruence. Théorème 1.2. Soit a b
Let n ? N and ab ? Z. We say that a is congruent to b modulo n
a divise b s'il existe un entier relatif k tel que b = ka. Deux entiers a et b sont congrus modulo n lorsque a – b est divisible par n.
We will say that two integers a and b are congruent modulo n and we write a ? b mod n If a ? b mod n
Integer a is congruent to integer b modulo m > 0 if a and b give If a ? b (mod m)
"a is congruent to b modulo m" means m
"a is congruent to b modulo m" means m
b)+(b ? a). D. We now show that congruence mod n plays well with the basic arithmetic ... ak ? bk mod n. Proof. This is just an easy induction on k.
(Hint: Most of the terms are congruent to 0.) (3) Show that if n