Divisibility Proof: Prove n^7 - n Divisible by 7

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Homework Help Overview

The discussion revolves around proving that for any positive integer \( n \), the expression \( n^7 - n \) is divisible by 7. Participants are exploring various methods to approach this proof, with a specific emphasis on breaking the problem into 7 cases based on the possible remainders when \( n \) is divided by 7.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the necessity of breaking the proof into 7 cases, questioning whether other methods, such as mathematical induction, could be acceptable. Some suggest using the remainders 0 through 6 and proof by exhaustion, while others reference Fermat's Little Theorem and modular arithmetic as potential avenues for exploration.

Discussion Status

The discussion is ongoing, with participants offering hints and alternative approaches. Some express uncertainty about how to draw conclusions from their attempts, indicating a lack of consensus on the best method to proceed.

Contextual Notes

Participants are required to adhere to the method of breaking the proof into 7 cases, which may limit the exploration of other proof techniques. There is also mention of factoring the original expression as part of the discussion.

tysonk
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Can someone help me with the following proof.
prove that if n is a positive integer then n^7 - n is divisible by 7. This should be done by breaking it down into 7 cases.
 
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welcome to pf!

hi tysonk! welcome to pf! :wink:

(try using the X2 icon just above the Reply box :smile:)
tysonk said:
Can someone help me with the following proof.
prove that if n is a positive integer then n^7 - n is divisible by 7. This should be done by breaking it down into 7 cases.

hint: what do you think the 7 cases might be? :wink:
 
tysonk said:
Can someone help me with the following proof.
prove that if n is a positive integer then n^7 - n is divisible by 7. This should be done by breaking it down into 7 cases.

n x 2n x 3n x 4n x ... x 6n == 6! (mod 7) in some permuatation.
6! x n^6 == 6! (mod 7)
n^7 / n == 1 (mod 7)

now you can draw conclusion.. or refer to generalised theorem- fermat's little theorem.
 
"Should be done" or "Must be done" by breaking into cases of 7's? Would proof using mathematical induction be acceptable for you?
 
It should be done using the 7 cases.
I know the 7 cases are the remainders. 0, 1, 2, 3, 4, 5, 6 and using proof by exhaustion for the 7 cases. (and perhaps factoring the original)
However, I'm still not able to draw a conclusion. The help is appreciated.
 
tysonk said:
It should be done using the 7 cases.
I know the 7 cases are the remainders. 0, 1, 2, 3, 4, 5, 6 and using proof by exhaustion for the 7 cases. (and perhaps factoring the original)
However, I'm still not able to draw a conclusion. The help is appreciated.

ok, take remainder 3, so n = 7k + 3 …

then n7 - n = (7k + 3)7 - (7k + 3) = … ? :wink:
 

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