- #1

- 26

- 0

## Main Question or Discussion Point

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

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

- Thread starter xax
- Start date

- #1

- 26

- 0

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

- #2

- 1,056

- 0

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.

Last edited:

- #3

- 26

- 0

Edit: p<>2 means p can't be 2.

- #4

- 1,056

- 0

Because the order is even. If a^t-1 = 0 Mod p^k, then the order is odd.

- #5

- 26

- 0

got it robert, thank you.

- Replies
- 2

- Views
- 7K

- Last Post

- Replies
- 1

- Views
- 3K

- Last Post

- Replies
- 4

- Views
- 6K

- Last Post

- Replies
- 7

- Views
- 8K

- Last Post

- Replies
- 4

- Views
- 7K

- Last Post

- Replies
- 3

- Views
- 6K

- Replies
- 1

- Views
- 2K

- Last Post

- Replies
- 9

- Views
- 3K

- Replies
- 5

- Views
- 3K

- Replies
- 1

- Views
- 2K