Mathematics – difficult and comprehensive science. Without knowing a formula, it is impossible to solve a simple problem of a subject. What to tell about such cases when the solution of a task requires something bigger, than just to remove one formula and to substitute the available values. Finding of an antiderivative from a root concerns those.
Instruction
1. It is worth specifying that here finding of a primitive root which on module n number g – it that all degrees of this number on module n are passed on all to numbers, mutually prime with n, is called means. Mathematically it can be expressed so: if g is a primitive root on module n, then for any integer, it that gcd (a, n) = 1, will be such number k that g^k ≡ a (mod n).
2. The theorem which shows that if the smallest number k for which g^k ≡ 1 (mod n), equals Φ(n) then of g is a primitive root was provided in the previous step. From here it is visible that k is g indicator. For any an Euler's theorem – a^ (Φ (n)) ≡ 1 (mod n) – therefore is carried out to check that g is a primitive root, it is enough to make sure that for all smaller Φ(n) numbers d g^d ≢ is carried out by 1 (mod n). However this algorithm quite slow.
3. From Lagrange's theorem it is possible to draw a conclusion that the indicator of any of numbers on module n is a divider Φ(n). It simplifies a task. It is enough to make sure that for all own dividers of d | Φ(n) g^d ≢ is carried out by 1 (mod n). This algorithm already much quicker than previous.
4. Factorize number Φ(n) = p_1^ (a_1) … p_s^(a_s). Prove that in the algorithm described in the previous step as d it is enough to consider only numbers of the following look: Φ(n)/p_i. Really, start up d - it is any own divider Φ(n). Then, obviously, there will be such j that d | Φ(n)/p_j, that is d*k = Φ(n)/p_j.
5. But if g^d ≡ 1 (mod n), then at us would leave g^ (Φ (n)/p_j) ≡ g^ (d * k) ≡ (g^d) ^k ≡ 1^k ≡ 1 (mod n). That is it turns out that among look numbers Φ(n)/p_j there would be it for which the condition would not be satisfied that, actually, and was required to be proved.
6. The algorithm of finding of a primitive root, thus, will look as follows. At first is Φ(n), then it is factorized. Later all numbers g = 1 … n get over, and for each of them all sizes Φ(n)/p_i are considered (mod n). In case for the current g all these numbers are other than unit, it is g and will be a required primitive root.
7. If to consider that the number Φ(n) has O (log Φ(n)), and exponentiation is carried out by means of an algorithm of binary exponentiation, that is for O (log n), it is possible to learn operating time of an algorithm. And it is equal to O (Ans * log Φ(n) * logn) + t. Here t is time of factorization of number Φ(n), and Ans is a result, that is value of a primitive root.