
#1
Mar1908, 01:14 AM

P: 26

p<>2, prime and ord p^k (a) = 2t.




#2
Mar2108, 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 (x1)(x+1) == 0 Mod P^k. So in an integral domain, one of the terms under discussion (a^t1) or (a^t+1) will be equal to zero, Mod p^k.




#3
Mar2308, 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^t1=0 mod p^k?
Edit: p<>2 means p can't be 2. 



#4
Mar2308, 01:54 PM

PF Gold
P: 1,059

Prove that a^t = 1 (mod p^k) for....
Because the order is even. If a^t1 = 0 Mod p^k, then the order is odd.




#5
Mar2308, 02:53 PM

P: 26

got it robert, thank you.



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