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

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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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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