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

by xax
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xax
#1
Mar19-08, 01:14 AM
P: 26
p<>2, prime and ord p^k (a) = 2t.
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robert Ihnot
#2
Mar21-08, 03:13 AM
PF Gold
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
PF Gold
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.


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