# [SOLVED] Proof of divisibility by Induction

• Nov 14th 2007, 04:26 PM
R.Laramee
[SOLVED] Proof of divisibility by Induction
I am having difficulty with induction in several areas, however one question at a time.
Ex. if 6 divides n(n^2 + 5) for all n >=1 prove for n+1.

What I have so far is (k+1)((k+1)^2 +5)

= k^3 + 5k + 3k^2 + 3k + 6

which I have worked out to

k(k^2 +5) + 3k^2 + 3k + 6

I know 6| k(k^2 +5) but how to I prove 6 divides the rest?
Am I able to use
3k^2 mod 6 + 3k mod 6 + 6 mod 6 = 3+3+0 mod 6 = 0 mod 6 therefore also divisible by 6?
• Nov 14th 2007, 07:40 PM
Soroban
Hello, R.Laramee!

Quote:

If 6 divides $n(n^2 + 5)$ for all $n \geq 1$, prove for $n+1.$

What I have so far is: . $(k+1)[(k+1)^2 +5] \:= \:k^3 + 5k + 3k^2 + 3k + 6$
. . which I have worked out to: . $k(k^2 +5) + 3k^2 + 3k + 6$

I know $6\,|\,k(k^2 +5)$, but how to I prove 6 divides the rest?

Here one approach . . .

Actually, we know that: . $k(k^2+5) {\color{blue}\,+\, 6}$ is divisible by 6.
. . So we are concerned with: . $3k^2 + 3k$

Note that: . $3k^2+3k \:=\:3\!\cdot\!k(k+1)$
. . This is: . $3 \times (\text{product of two consecutive integers})$

With two consecutive integers, one must be even and the other must be odd.

Hence, $k(k+1)$ is even . . . Therefore, $3k(k+1)$ is a multiple of 6.

• Nov 14th 2007, 08:15 PM
R.Laramee
That is a really cool way of looking at it. Thanks