Thread: [SOLVED] Proof of divisibility by Induction

1. [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?

2. Hello, R.Laramee!

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

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

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

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

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

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

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

3. That is a really cool way of looking at it. Thanks