Solving Arithmetics Problem: Find m Number for a^(m-1) ≡ 1 mod m

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Homework Help Overview

The discussion revolves around proving the existence of infinitely many composite numbers \( m \) such that \( a^{m-1} \equiv 1 \mod m \), where \( a \) is a non-zero integer. Participants are exploring the conditions under which this congruence holds and the nature of the numbers involved.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to identify specific composite numbers \( m \) that satisfy the given condition. Some question the original poster's ability to find even one such number, while others suggest testing various values of \( m \). There is also mention of trivial solutions and specific pairs of \( a \) and \( m \) that have been found.

Discussion Status

The discussion is ongoing, with some participants providing examples of potential solutions and others questioning the assumptions made by the original poster. There are indications of differing interpretations of the problem, particularly regarding the nature of \( m \) as composite rather than prime.

Contextual Notes

There is a noted distinction between composite and prime numbers in the context of the problem, which has led to some confusion. Additionally, the density of the solutions being discussed, particularly in relation to Fermat pseudoprimes, is acknowledged.

luciasiti
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Hi everyone
I'm having difficulty in proving the following problem
Give a\in \mathbb{Z} and a \ne 0. Prove that: there exist infinitely many composite numbers m such that a^{m-1}\equiv 1 mod m
I tried to find m number, which is satisfied with problem requirement, but I still haven't found it.
Can you help me find m number?
Thanks for your help!
 
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You are trying to prove there exist infinitely many and you cannot find even 1? You need to talk to your teacher about this!

Suppose a= 3. Then you want to find prime m such that am-1= 1 (mod m).

Okay, just try some prime numbers. If m= 5 then am-1= 34= 81= 1 (mod 5). Done!

Perhaps I am misunderstanding your question.
 
HallsofIvy said:
You are trying to prove there exist infinitely many and you cannot find even 1? You need to talk to your teacher about this!

Suppose a= 3. Then you want to find prime m such that am-1= 1 (mod m).

Okay, just try some prime numbers. If m= 5 then am-1= 34= 81= 1 (mod 5). Done!

Perhaps I am misunderstanding your question.

Thank you. But, in this problem, m is not prime number, m is composite number and a is an any integer.
 
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If m is odd then a=m-1 is a trivial solution. Other solutions are quite sparse. In the form a;m I found
4;15, 4;85, 4;9, 15;85, 7;77, 8;21, 8;45, 8;63, 8;105, 8;65, 11;15, 11;45, 15;49, 16;51, 16;85, 16;91, 18;115, 19;105, 20;85, 21;69, 27;33, 31;33, 31;77, 32;93, 34;77, 35;99, 37;55, 38;115
 
haruspex said:
If m is odd then a=m-1 is a trivial solution. Other solutions are quite sparse. In the form a;m I found
4;15, 4;85, 4;9, 15;85, 7;77, 8;21, 8;45, 8;63, 8;105, 8;65, 11;15, 11;45, 15;49, 16;51, 16;85, 16;91, 18;115, 19;105, 20;85, 21;69, 27;33, 31;33, 31;77, 32;93, 34;77, 35;99, 37;55, 38;115
Typo: should have been 4;91, not 4;9.
Have now found a general solution for a even: set m = a2-1. Might be extensible to an infinite set of m values somehow.
 
The m values are the Fermat pseudoprimes to base a. There are an infinity of them but they are generally not very dense; much less dense than primes.

A proof is here: http://primes.utm.edu/notes/proofs/a_pseudoprimes.html - you aren't going to stumble on this by trial and error.
 

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