Register to reply

Prove that a^t = -1 (mod p^k) for...

by xax
Tags: prove
Share this thread:
xax
#1
Mar19-08, 01:14 AM
P: 26
p<>2, prime and ord p^k (a) = 2t.
Phys.Org News Partner Science news on Phys.org
'Office life' of bacteria may be their weak spot
Lunar explorers will walk at higher speeds than thought
Philips introduces BlueTouch, PulseRelief control for pain relief
robert Ihnot
#2
Mar21-08, 03:13 AM
P: 1,059
I am not certain about the meaning of p<>2, but if the order of a is 2t, then a^t==-1 Mod p^k....Because there are only two elements of order 2, and phi(2) =1, which tells us only one element belongs to 2. That is, if x^2==1 Mod P^k, then (x-1)(x+1) == 0 Mod P^k. So in an integral domain, one of the terms under discussion (a^t-1) or (a^t+1) will be equal to zero, Mod p^k.
xax
#3
Mar23-08, 10:25 AM
P: 26
Thanks robert for your help, but how can I say that only a^t+1 = 0 mod p^k and it's not possible a^t-1=0 mod p^k?
Edit: p<>2 means p can't be 2.

robert Ihnot
#4
Mar23-08, 01:54 PM
P: 1,059
Prove that a^t = -1 (mod p^k) for...

Because the order is even. If a^t-1 = 0 Mod p^k, then the order is odd.
xax
#5
Mar23-08, 02:53 PM
P: 26
got it robert, thank you.


Register to reply

Related Discussions
Please prove this Differential Geometry 6
Need help proving an expression of roots of sums including roots General Math 16
To Prove Calculus 3
Prove the following: Introductory Physics Homework 3
Is this ok to prove? Introductory Physics Homework 6