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

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Discussion Overview

The discussion revolves around proving the statement that \( a^t \equiv -1 \mod p^k \) under the condition that \( p \) is a prime not equal to 2 and the order of \( a \) modulo \( p^k \) is \( 2t \). The focus includes mathematical reasoning and exploration of properties related to orders of elements in modular arithmetic.

Discussion Character

  • Mathematical reasoning, Debate/contested

Main Points Raised

  • Some participants assert that if the order of \( a \) is \( 2t \), then \( a^t \equiv -1 \mod p^k \) follows from the properties of elements of order 2.
  • One participant expresses uncertainty about the meaning of \( p \neq 2 \) and discusses the implications of the order of \( a \) being \( 2t \), suggesting that if \( x^2 \equiv 1 \mod p^k \), then either \( x-1 \equiv 0 \) or \( x+1 \equiv 0 \mod p^k \).
  • Another participant questions how to conclude that \( a^t + 1 \equiv 0 \mod p^k \) is the only possibility, while also noting that \( p \neq 2 \) clarifies the context.
  • It is noted that if \( a^t - 1 \equiv 0 \mod p^k \), then the order would be odd, which contradicts the given condition of \( 2t \) being even.

Areas of Agreement / Disagreement

Participants express differing views on the implications of the order of \( a \) and the conditions under which \( a^t \equiv -1 \mod p^k \) holds. The discussion remains unresolved regarding the specific reasoning behind the exclusivity of \( a^t + 1 \equiv 0 \mod p^k \).

Contextual Notes

There are limitations in the discussion regarding the assumptions made about the order of elements and the implications of modular arithmetic properties, which are not fully explored or resolved.

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

p<>2, prime and ord p^k (a) = 2t.
 
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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:
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.
 
Because the order is even. If a^t-1 = 0 Mod p^k, then the order is odd.
 
got it robert, thank you.
 

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