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

[tysonk]

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.

 

[tiny-tim]

welcome to pf!

hi tysonk! welcome to pf!

(try using the X² icon just above the Reply box )

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?

 

[icystrike]

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.

 

[symbolipoint]

"Should be done" or "Must be done" by breaking into cases of 7's? Would proof using mathematical induction be acceptable for you?

 

[tysonk]

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.

 

[tiny-tim]

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) = … ?