Solving Upper Congruences: Step-by-Step Guide for x^{n}+a=mod(b)

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Discussion Overview

The discussion revolves around solving the upper congruence equation \( x^{n} + a \equiv b \mod(c) \) for integer values of \( a, b, n \) where \( n > 0 \). Participants explore methods for solving this type of congruence, including numerical and approximate methods, and compare it to simpler linear congruences.

Discussion Character

  • Exploratory
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant seeks a step-by-step example for solving \( x^{n} + a \equiv b \mod(c) \) after having solved linear congruences.
  • Another participant expresses confusion about the initial question and suggests that reciprocity might be applicable depending on the interpretation.
  • A participant mentions knowing how to solve for \( n=1 \) and inquires about numerical methods for finding solutions to the general case.
  • One participant argues that if roots can be easily calculated for \( x^{n} - a \equiv 0 \), it would undermine the security of RSA, implying that such calculations are inherently difficult.
  • Another participant challenges the simplicity of solving the congruence, suggesting that understanding functions and algorithms is crucial for addressing the problem effectively.
  • A participant proposes a specific function and method for finding roots related to the congruence, involving the floor function and a derived equation.
  • One participant challenges the practicality of the proposed methods, questioning their efficiency compared to exhaustive search methods.

Areas of Agreement / Disagreement

Participants express differing views on the feasibility and methods for solving upper congruences, with no consensus reached on a specific approach or solution. The discussion remains unresolved with multiple competing ideas presented.

Contextual Notes

Participants reference various mathematical concepts and methods, but there are limitations in the clarity of definitions and assumptions regarding the congruence and the proposed solutions. The discussion does not resolve these ambiguities.

lokofer
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If a,b,n are integers..how could you solve:

[tex]x^{n}+a=mod(b)[/tex] n>0 and integer..

If possible please put an "step by step" example..i have managed to solve the linear congruence

[tex]ax=bmod(c)[/tex] but i don't know how to solve "upper" congruences..thanks.
 
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I really don't know what you're asking. I presume the question you've already solved is [tex]ax\equiv b\pmod c[/tex] but I'm not sure what your first is. Depending on what you mean, you might be able to use reciprocity.
 
Last edited:
i know how to solve for n=1 then i would like to know how to solve the special case y proposed...

- By the way...are not there "approximate" or numerical method to calculate the set of the solutions?..
 
If you can solve for any x other than 1 (or 2) easily then you've just proven RSA easy to break. Come on, Jose, think for a second before posting a question: taking roots mod c is hard, and that's the whole point of RSA.
 
- I don't know why is the "congruence" [tex]x^{n}=amod(b)[/tex] for n>1 in fact it's nothing but a Monomial..you could calculate all the roots of [tex]x^{n}-a=0[/tex] easily and from this find a solution... :rolleyes: :rolleyes:
 
Then why don't you 'just do it'? You really ought to get straight in your mind what a function is, what an algorithm to find a value of said function is, and what you consider to be a reasonable cost for carrying out the algorithm, and then state what that is.
 
lokofer said:
i know how to solve for n=1 then i would like to know how to solve the special case y proposed...

- By the way...are not there "approximate" or numerical method to calculate the set of the solutions?..
What would be an "approximation" to an integer?
 
-For example "Hallfosftivy"..a solution to the (general ) congruence:

[tex]f(x)=amod(x)[/tex] (for example) is the solution to the "equation" (non-linear) in the form:

[tex][\frac{f(x)-a}{x}]-\frac{f(x)-a}{x}=0[/tex] or if we define the function:

[tex]<x>=x-[x][/tex] where "[x] " is the floor function then:

[tex]<\frac{f(x)-a}{x}>=0[/tex] so from this you could calculate all the "roots" (integers) and solution to the initial congruence...
 
Go on then. Do it. Show that it's better than a simple exhastive search through phi(c) options, (or c-1 if you don't know the factorization of c).
 

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