What can a possible equation be for the multiplicative order be?

Question

I'm taking Discrete & I know that a^k = 1 mod n, but is there a shortcut to finding what k equals??

if u can help plzz thnks :]

Answer 1

There are only a few rules that help narrow down the order.
If n is a prime p then a^(p-1) = 1(mod p). This is Fermat's
little theorem. (Note that a must not be a multiple of p.)
So the order of a mod n divides p-1. As to
which divisor you want, you simply have to try them all.
Next, suppose that a and n are relatively prime.
Let φ(n) be the number of integers less than n
and relatively prime to n. (Euler's phi function)
Then a^φ(n) = 1(mod n) and the order of a mod n
divides φ(n). Again, we don't know which divisor
of φ(n) works. You just have to try them all.
Examples:
10 has order 6 mod 7, 10 has order 2 mod 11
10 has order 6 mod 13 but 10 has order 16 mod 17.
φ(24) = 8, but all possible a (1,5,7,11,13,17,19
and 23) have order 2.
It's still an open question whether there
are infinitely many primes p for which
10 has order p-1 mod p.
Hope that helps!