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

In summary, The conversation discusses how to prove that if n is a positive integer then n^7 - n is divisible by 7 by breaking it down into 7 cases. The conversation also mentions using proof by exhaustion and factoring the original equation. One participant suggests using mathematical induction while another suggests considering the remainders of 7 as the 7 cases. The conversation concludes with a hint about taking remainder 3 and using it to find the solution.
  • #1
tysonk
33
0
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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  • #2
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:
 
  • #3
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.
 
  • #4
"Should be done" or "Must be done" by breaking into cases of 7's? Would proof using mathematical induction be acceptable for you?
 
  • #5
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.
 
  • #6
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:
 

1. What is a divisibility proof?

A divisibility proof is a mathematical method used to show that one number is divisible by another. It involves using logical steps and mathematical principles to demonstrate that the quotient of the two numbers is a whole number, with no remainder.

2. How do you prove n^7 - n is divisible by 7?

To prove that n^7 - n is divisible by 7, we can use the principle of mathematical induction. First, we show that the statement is true for n=1. Then, we assume it is true for some integer k, and use this assumption to prove that it is also true for k+1. This establishes that the statement is true for all positive integers, and therefore n^7 - n is divisible by 7.

3. Why is proving n^7 - n divisible by 7 important?

Proving that n^7 - n is divisible by 7 is important as it is a useful tool in many other mathematical proofs and applications. It also helps to strengthen our understanding of number theory and algebraic concepts.

4. Are there any other divisibility proofs for different numbers?

Yes, there are many different divisibility proofs for different numbers. Some commonly used proofs include proving divisibility by 2, 3, 5, 9, and 11. Each proof may use different mathematical principles and may require different steps to be demonstrated.

5. Can n^7 - n be divisible by a number other than 7?

Yes, n^7 - n can be divisible by numbers other than 7. However, the question specifically asks for a proof that n^7 - n is divisible by 7, which means that we are focusing on proving this specific statement rather than exploring other potential divisors.

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