# Thread: 6 is a factor of n^(3)-n

1. ## 6 is a factor of n^(3)-n

Use mathematical induction to prove that the given statement is true for all positive integers n.

can some one teach me how to start the proof and how to end it? Thank you.

2. For induction, there are 2 steps.
1. Show that some concrete value satisfies your relation.
2. Suppose the relation is true for n, and then show it is true for n+1.

For your case, you can show that
6 | 1^3 - 1.

Then assume that
6 | n^3-n for some n , and show that
6 | (n+1)^3-(n+1), where this n is the same n in the previous line.

Hint:
Expand the cubic term, note that the terms remaining after the subtraction are 3(n)(n+1) and show that
6 | 3(n)(n+1)

3. Originally Posted by qmech
For induction, there are 2 steps.
1. Show that some concrete value satisfies your relation.
2. Suppose the relation is true for n, and then show it is true for n+1.

For your case, you can show that
6 | 1^3 - 1.

Then assume that
6 | n^3-n for some n , and show that
6 | (n+1)^3-(n+1), where this n is the same n in the previous line.

Hint:
Expand the cubic term, note that the terms remaining after the subtraction are 3(n)(n+1) and show that
6 | 3(n)(n+1)

I worked it out toward the end where I got stuck: 6| (n) (n+1) (n+2)

and then, i do not know what i should do with it. It doesn't seem to have any relationship with the 6...

4. If you are going to use induction, I don't think it is a good idea to factor $\displaystyle n^3- n$.

If 6 divides $\displaystyle n^3- n$, then $\displaystyle n^3- n= 6m$ for some integer m. Then $\displaystyle (n+1)^3- (n+1)= n^3+ 3n^2+ 3n+ 1- n- 1$$\displaystyle = n^3- n+ 3n^2+ 3n= 6m+ 3n(n+1)$
Now do two cases:
1: n is even: n= 2k
2: n is odd: n= 2k+1, n+1= 2k+2= 2(k+1).
In either case, n(n+1) is even so 3n(n+1) is also divisible by 6.

But you don't really need to use induction.

$\displaystyle n^3- n= n(n-1)(n+1)$ is the product of three consecutive integers. Certainly at least one must be even and one of them must be a multiple of 3.