Is There a Solution to the Equation A=x^b mod p?

  • Context: Undergrad 
  • Thread starter Thread starter Chu
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Discussion Overview

The discussion revolves around the equation \( a = x^b \mod p \), where \( p \), \( a \), and \( b \) are known, and seeks to explore potential methods for solving this equation. The context includes aspects of cryptography and modular arithmetic.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested

Main Points Raised

  • Chu expresses uncertainty about solving the equation \( a = x^b \mod p \) and believes a solution exists.
  • One participant suggests trying to solve a specific case, \( x^2 = 2 \mod 3 \), as a potential example.
  • Another participant clarifies that they believe a solution exists given the context of a cryptographic algorithm, where \( x \) is known but not \( a \) or \( b \).
  • A different viewpoint proposes that while it is possible to solve the equation by raising every remainder to the \( b \)th power, this approach may not yield a better solution in general cases.
  • This participant also notes that \( b \) must divide \( \varphi(x) \), but questions whether this condition is sufficient for finding a solution or determining its nature.
  • Concerns are raised about the complexity of finding generators of the group of units mod a prime, indicating the difficulty of the problem at hand.

Areas of Agreement / Disagreement

Participants express differing views on the solvability of the equation and the methods to approach it. There is no consensus on a definitive solution or the conditions required for one.

Contextual Notes

Participants mention the need for specific conditions, such as \( b \) dividing \( \varphi(x) \), but do not resolve the implications of these conditions on the existence of solutions.

Chu
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Say you are given a=x^b mod p, where p, a, and b are known. Is there a way to solve this? I am pretty sure there is . . . but it is driving me nuts.

-Chu
 
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Try solving x^2 = 2 (mod 3).
 
Muzza said:
Try solving x^2 = 2 (mod 3).

Sorry, I'll rephrase. I know a solution must exist from the choices of p,b,and a (this is part of a crypto algorithm where they know x, I do not, and I am wondering if I have sufficent info to solve for it).
 
You can of course solve it - all one needs is to raise every remainder to the b'th power. However, I don't think there is a better solution except in certain circumstances.

obviously b must divide [tex]\varphi(x)[/tex], but that doesn't give a sufficient condition for a solution, or even tell you what it is.

The reason for my scepticism is that it is a well known and difficult problem to find generators of the group of units mod a prime (which is a cyclic group).
 

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