Prove by induction that r(r-1)(r+1) is an even integer

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1. Aug 13, 2016

sooyong94

1. The problem statement, all variables and given/known data
Prove by induction, that when r(r-1)(r+1) is an even integer when r=2,3,4....

2. Relevant equations
Prove by induction

3. The attempt at a solution
I began with the base case r=2, leading 6.
Then I proceed with r=3, leading 24.

Now if r=k is true, then k(k-1)(k+1) is also true.
If r=k+1, then (k+1)(k)(k+2)
But now I'm stuck at this point - how do I proceed with this?

2. Aug 13, 2016

blue_leaf77

Write k(k-1)(k+1)=2F where F is an integer.

3. Aug 13, 2016

sooyong94

k(k-1)(k+1) = 2F
k^3 - k=2F

k(k+1)(k+2) = k^3 +3k^2 +2k
= k^3 - k +3k^2 +3k
=2F+3k(k+1)

4. Aug 13, 2016

blue_leaf77

You can still substitute k(k+1) using k(k-1)(k+1)=2F.

5. Aug 13, 2016

sooyong94

2F+3(2F/(k-1) ?

6. Aug 13, 2016

blue_leaf77

Yes. Now you should see why it is an even integer.

7. Aug 13, 2016

sooyong94

Ah I see already - since 2F= 2F+6F/(k-1), when k is not equal to 1, it is divisible by 2.

8. Aug 13, 2016

blue_leaf77

I don't know if that's a typo or on purpose, but the LHS and RHS cannot be equal. Moreover, apart from being divisible by two you should also be convinced that 2F+6F/(k-1) is indeed an integer.

9. Aug 13, 2016

sooyong94

10. Aug 13, 2016

Staff: Mentor

Somewhat OT, but in fact, (r - 1)r(r + 1) is divisible by 6, for r = 1, 2, 3, ...

11. Aug 14, 2016

sooyong94

Looks like I managed to work them out. Please mark this thread as solved. ;)

12. Sep 2, 2016

micromass

Staff Emeritus
You know, induction isn't really needed here. For any $r$, either $r$ or $r+1$ is even. Multiplying any integer by an even number yields an even number.