# Thread: minimal power and congruence

1. ## minimal power and congruence

Hi,

------------------------------------
Problem:
Let a and b be integers.
Let p be an odd prime divisor of a^16 + 1. (i.e. a^16 == -1 mod p)
Show that if e is the smallest positive integer such that b^e == 1 mod p,
then e=32.
-------------------------------------

This is kind of a weird problem I know. Just using the fact that p is an odd prime divisor of a^16 + 1 I showed that p == 1 mod 32. That might not be useful, but the number 32 is in there so it might be relevant.

Also, I know that b^x == 1 mod p if and only if e divides x, with x>0.

Another thing I know is that if b^x == 1 == b^y mod p, with x and y positive, then a^d == 1 mod p where d = gcd(x,y).

I'm not sure what is relevant and what is not, I've been messing with it for a while. Does anyone have any hints? If you need the problem clarified more let me know. Thanks.

2. ## Clarity

Just to clarify, your conjecture is for natural number a and odd prime p satisfying $\displaystyle a^{16}\equiv 1(\bmod p)$, there are no solutions to $\displaystyle b^e\equiv -1 (\bmod p)$ for $\displaystyle e<32$. Is this correct?