# Thread: Divisibility by 6 Induction

1. ## Divisibility by 6 Induction

Hey all, some help finishing this proof would be appreciated:

For all k in the Naturals, k^3 + 5k is divisible by 6.

Proof:
1^3 + 5(1) = 6 which is divisible by 6

P(n): n^3 +5n = 6y

(n+1)^3 + 5(n+1) = 6z
(n^3 + 3n^2 + 3n + 1) + 5(n+1)
6y + 3n^2 + 3n + 6 = 6z
3(2y + n^2 + n + 2) = (3*2)z

where do we go from here? How can I show divisibility by 6 here? Thanks in advance for the help!

2. Originally Posted by jstarks44444
Hey all, some help finishing this proof would be appreciated:

For all k in the Naturals, k^3 + 5k is divisible by 6.

Proof:
1^3 + 5(1) = 6 which is divisible by 6

P(n): n^3 +5n = 6y

(n+1)^3 + 5(n+1) = 6z
(n^3 + 3n^2 + 3n + 1) + 5(n+1)
6y + 3n^2 + 3n + 6 = 6z

where do we go from here? How can I show divisibility by 6 here? Thanks in advance for the help!
I deleted your last line above.

6y+6 is divisible by 6

$3n^2+3n=3n(n+1)$

One of n and n+1 is even.