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Integer multiplication respects the congruence classes, that is, a ≡ a' and b ≡ b' (mod n) implies ab ≡ a'b' (mod n). This implies that the multiplication is associative, commutative, and that the class of 1 is the unique multiplicative identity. Finally, given a, the multiplicative inverse of a modulo n is an integer x satisfying ax ≡ ...
If p is a prime number which is not a divisor of b, then ab p−1 mod p = a mod p, due to Fermat's little theorem. Inverse: [(−a mod n) + (a mod n)] mod n = 0. b −1 mod n denotes the modular multiplicative inverse, which is defined if and only if b and n are relatively prime, which is the case when the left hand side is defined: [(b −1 ...
A prime modulus requires the computation of a double-width product and an explicit reduction step. If a modulus just less than a power of 2 is used (the Mersenne primes 2 31 − 1 and 2 61 − 1 are popular, as are 2 32 − 5 and 2 64 − 59), reduction modulo m = 2 e − d can be implemented more cheaply than a general double-width division ...
Even without knowledge that we are working in the multiplicative group of integers modulo n, we can show that a actually has an order by noting that the powers of a can only take a finite number of different values modulo n, so according to the pigeonhole principle there must be two powers, say s and t and without loss of generality s > t, such that a s ≡ a t (mod n).
A prime number (or prime) is a natural number greater than 1 that has no positive divisors other than 1 and itself. ... (mod p 2): 2, 20771, 40487, 53471161, ...
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The base step, that 0 p ≡ 0 (mod p), is trivial. Next, we must show that if the theorem is true for a = k, then it is also true for a = k + 1. For this inductive step, we need the following lemma. Lemma. For any integers x and y and for any prime p, (x + y) p ≡ x p + y p (mod p). The lemma is a case of the freshman's dream. Leaving the ...
This result may be deduced from Fermat's little theorem by the fact that, if p is an odd prime, then the integers modulo p form a finite field, in which 1 modulo p has exactly two square roots, 1 and −1 modulo p. Note that a d ≡ 1 (mod p) holds trivially for a ≡ 1 (mod p), because the congruence relation is compatible with exponentiation.