Intro to number theory congruence problem 1

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Homework Help Overview

The problem involves demonstrating that for a composite number \( n > 4 \), \( n \) divides \( (n-1)! \) and consequently that \( (n-1)! \) is not congruent to -1 modulo \( n \). The discussion references Wilson's theorem and its implications for primality testing.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the reasoning behind why both factors of a composite number \( n \) must be present in \( (n-1)! \). There is an exploration of different cases regarding the equality of the factors \( a \) and \( b \) and their implications for the proof.

Discussion Status

Some participants have provided insights into the reasoning process, while others question the necessity of considering different cases separately. The discussion reflects a mix of interpretations and attempts to clarify the implications of the assumptions made in the problem.

Contextual Notes

There is a note on the professor's expectations for thoroughness in considering all possibilities, which adds a layer of complexity to the discussion regarding the treatment of cases where \( a = b \) versus \( a \neq b \).

Proggy99
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Homework Statement


If n>4 is a composite number, show that n|(n-1)! Conclude that (n-1)! not congruent -1(mod n).

(This shows that Wilson's theorem can be used as a proof of primality. It is unfortunately not practical for large numbers)

Homework Equations



The Attempt at a Solution


I know in words why a composite (ab) number does not work, but am not really sure how to prove it. Can anyone give me a jumping off point for this problem please?
 
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If n= ab, then clearly a< n and b< n so both are factors in (n-1)!.
 
HallsofIvy said:
If n= ab, then clearly a< n and b< n so both are factors in (n-1)!.

and if a|(n-1)! and b|(n-1)!, then ab|(n-1)! and since n=ab, then n|(n-1)!

Then since (n-1)! congruent 0(mod n), then (n-1)! not congruent -1(mod n)

Did I follow that through correctly?

It strikes me that if a=b, then this might break down my reasoning and the problem did not specify whether a=b or not or if it mattered. I will have to think on that a little.
 
found the answer to my a=b or n=a^2 question. Seems that there are three possibilities to consider...when a not = b, when a=b and a is prime, when a=b and a is not prime. When a not = b and when a=b and a is not prime are easier to prove. a=b and a is prime relies on the additional information that n>4 to prove. Thanks for the help on this problem!
 
I don't see why those have to be considered separately. If n= ab, why does it matter if a= b or not?
 
HallsofIvy said:
I don't see why those have to be considered separately. If n= ab, why does it matter if a= b or not?


well, my professor expects us to be pretty thorough when considering all the possibilities and explaining them away. For instance:

if n=9=3*3, then ab=3*3 and (n-1)! = 1*2*3*4*5*6*7*8
The reason it still works for 3*3 is because both 3 and 2*3 occur in the problem, and in fact when n=c^2, both c and 2*c will always occur in the problem. This is because
p=n(1/2) <= (n-1)/2 for n>4, thus 2p<= n-1

I would imagine that my professor would have marked me down for not taking that into consideration, as he has done on other problems in the past. I am curious on whether you would consider it necessary to include the additional explanation or if you consider it part of the shorter explanation? Thanks again for the help HoI.
 

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