Proving: 6 Divides (n^3-n) for All Integers n

  • Thread starter chan8366
  • Start date
  • Tags
    Integers
In summary, to prove that 6 divides (n^3-n) for all integers n, the remainder on division by 3 of the three factors (n)(n+1)(n-1) must be zero, which can be shown by considering cases where n=2k or n=2k+1 for some integer k. This is because one of any two consecutive numbers is even and one of any three consecutive numbers is a multiple of 3.
  • #1
chan8366
3
0

Homework Statement



prove:6 divides (n^3-n) for all integers n.

Homework Equations


n^3-n=(n)(n+1)(n-1)


The Attempt at a Solution


tried to use direct proof. Then used cases that involed n=2k for some integer k and n=2k+1 for some integer k. However, i could not get it so that 6 was factored from either odd/even of n^3-n i.e. n^3-n=6m for some integer m.
Just a hint please.
 
Physics news on Phys.org
  • #2
What's the remainder on division by 3 of the 3 factors you have exhibited?
 
  • #3
One of any two consecutive number is even. One of any three consecutive numbers is a multiple of 3.
 

Related to Proving: 6 Divides (n^3-n) for All Integers n

1. How can you prove that 6 divides (n^3-n) for all integers n?

To prove this statement, we can use mathematical induction. This means we will show that the statement holds true for the smallest integer value of n, and then prove that if it holds true for a particular value of n, it also holds true for the next value of n. This process will continue until we have shown that the statement holds true for all integers n.

2. What is the base case for the mathematical induction proof?

The base case for this proof is when n=1. We can substitute this value into the given statement, (1^3-1), which simplifies to 0. This shows that 6 does divide 0, as 0 divided by any number is equal to 0.

3. How do you prove that the statement holds true for the next value of n?

To prove that the statement holds true for the next value of n, we will assume that the statement is true for a particular value of n, and then use this assumption to prove that the statement is also true for the next value of n, which is n+1. This process is known as the inductive step.

4. How do you show that the statement holds true for all integers n?

By completing the base case and inductive step, we have shown that the statement holds true for n=1 and that if it holds true for a particular value of n, it also holds true for the next value of n. Therefore, the statement holds true for all integers n by mathematical induction.

5. Are there any other methods to prove that 6 divides (n^3-n) for all integers n?

Yes, there are other methods such as using the division algorithm or modular arithmetic. However, mathematical induction is the most commonly used method for proving statements like this one.

Similar threads

  • Calculus and Beyond Homework Help
Replies
7
Views
2K
  • Calculus and Beyond Homework Help
Replies
30
Views
2K
  • Calculus and Beyond Homework Help
Replies
12
Views
3K
  • Calculus and Beyond Homework Help
Replies
13
Views
2K
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
20
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
864
  • Calculus and Beyond Homework Help
Replies
3
Views
585
  • Calculus and Beyond Homework Help
Replies
5
Views
2K
  • Calculus and Beyond Homework Help
Replies
1
Views
561
Back
Top