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Fermat's little theorem: If p is prime and does not divide a, then a p−1 ≡ 1 (mod p). Euler's theorem: If a and m are coprime, then a φ(m) ≡ 1 (mod m), where φ is Euler's totient function. A simple consequence of Fermat's little theorem is that if p is prime, then a −1 ≡ a p−2 (mod p) is the multiplicative inverse of 0 < a < p.
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).
where ε denotes the signature of a permutation and π a is the permutation of the nonzero residue classes mod p induced by multiplication by a. For example, take a = 2 and p = 7. The nonzero squares mod 7 are 1, 2, and 4, so (2|7) = 1 and (6|7) = −1.
The set {3,19} generates the group, which means that every element of (/) is of the form 3 a × 19 b (where a is 0, 1, 2, or 3, because the element 3 has order 4, and similarly b is 0 or 1, because the element 19 has order 2). Smallest primitive root mod n are (0 if no root exists)
[1] For example, the expression "5 mod 2" evaluates to 1, because 5 divided by 2 has a quotient of 2 and a remainder of 1, while "9 mod 3" would evaluate to 0, because 9 divided by 3 has a quotient of 3 and a remainder of 0. Although typically performed with a and n both being integers, many computing systems now allow other types of numeric ...
X ≡ 6 (mod 11) has common solutions since 5,7 and 11 are pairwise coprime. A solution is given by X = t 1 (7 × 11) × 4 + t 2 (5 × 11) × 4 + t 3 (5 × 7) × 6. where t 1 = 3 is the modular multiplicative inverse of 7 × 11 (mod 5), t 2 = 6 is the modular multiplicative inverse of 5 × 11 (mod 7) and t 3 = 6 is the modular multiplicative ...
The number 3 is a primitive root modulo 7 [5] because = = = = = = = = = = = = (). Here we see that the period of 3 k modulo 7 is 6. The remainders in the period, which are 3, 2, 6, 4, 5, 1, form a rearrangement of all nonzero remainders modulo 7, implying that 3 is indeed a primitive root modulo 7.
If p is an odd prime and p − 1 = 2 s d with s > 0 and d odd > 0, then for every a coprime to p, either a d ≡ 1 (mod p) or there exists r such that 0 ≤ r < s and a 2 r d ≡ −1 (mod p). 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 ...