Thread: Just about done with this problem, but I am stuck on this last part!!

Q: Prove that 6 divides n^(3) - n whenever n is a nonnegative integer?
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What I have is this:

Basis Step:
If n = 0, then f(n)= n^3 - n = 0^(3) - 0 = 0 - 0 = 0.

So the basis step is true since its divisible by 6.
Inductive Hypothesis:
We assume true, if n = k, then f(k) = k^(3) - k.

We must prove that if n = k + 1, then f(k+1) = (k+1)^(3) – (k+1)
Inductive Step:
f(k+1) = (k+1)^(3) – (k+1)
= (k + 1)(k + 2)^(2) - (k + 1)
= (k+1)(k^(2) + 2k + 1) – (k+1)
= k^(3) + 2k^(2) + k + k^(2) + 2k + 1 – k – 1
= k^(3) + 3k^(2) + 3k + 1 – k – 1
= k^(3) + 3k^(2) + 3k – k
= (k^(3) - k) + (3k^(2) + 3k)
= (k^(3) - k) + 3(k^(2) + k)
= f(k) + 3(k^(2) + k)


Where I am stuck at is I know f(k) is divisible by 6 but the 3(k^(2) + k) is divisible by 3?

Hello, Grillakis!

Your work is excellent!

Prove that 6 divides for a nonnegative integer


What I have is this:

Basis Step:
If , then
. .
So the basis step is true since its divisible by 6.

Inductive Hypothesis:
We assume true: if , then: is divisible by 6.

We must prove that if ,
. . then: . is divisible by 6.


Inductive Step:

. . . . . .
. . . . . .
. . . . . .


Where I am stuck: is divisible by 6? . . . . Yes!

We know that is divisible by 6: .




Note that is the product of two consecutive integers.
Hence, one of them is even and the other is odd.
And their product is even, say,

So we have: .

Originally Posted by Soroban

Hello, Grillakis!

Your work is excellent!


We know that is divisible by 6: .




Note that is the product of two consecutive integers.
Hence, one of them is even and the other is odd.
And their product is even, say,

So we have: .

Thanks Soroban....I understand that now.