# Proving divisibility by induction

1. Sep 6, 2006

### leduc

Hello, I'm struggling with the question on induction.
I was wondering if you could help me?

Prove that n(n^2 +5) is divisible by 6 for n belonging to Z^+

P_1 is (1(1^2 + 5))/6=1 hence P_1 is true

If P_k is true then (k(K^2 +5))/6=r and if and only if (k(k^2 +5))=6r
then P_(k+1) is

(k+1)((K+1)^2 +5))=6r

We're looking for something with 6 as a factor.

2. Sep 6, 2006

### e(ho0n3

What have you done so far?

3. Sep 6, 2006

### HallsofIvy

Staff Emeritus
Yes, that's true.
No, of course not. you just said 6r was equal to K(K^2+ 1). (K+1)^2= K^2+ 2k+ 1 so (K+1)((K+1)^2+ 5)= (K+1)(K^2+ 1+ 2k+ 5)= K(K^2+ 1)+ what?

4. Sep 6, 2006

### 0rthodontist

No- you want (k+1)((k+1)^2 +5))=6s for some s. You've already used r in P_k, you can't use it again to mean something different in P_(k+1). Also be careful about capital letters--if you define k as lowercase it should be lowercase throughout.

Also, "If P_k is true then (k(K^2 +5))/6=r and if and only if (k(k^2 +5))=6r" is true whether or not P_k is true--it's just algebra and doesn't help the induction. You want to start with P_k and then derive P_(k+1).