Proof of divisibility of 3 sequential cubic numbers by induction

Question

The task is to prove that the sum of three subsequent cubic numbers is divisible by 9. My approach looks like this:

1. $n^3+(n+1)^3+(n+2)^3 = 9M$

2. Proof that the equation is true for $n=1$

   $1^3+2^3+3^3 = 1+8+27 = 36;$

   $ 36/9 = 4$

3. Assume that the equation is true for $n<=k$

   $k^3+(k+1)^3+(k+2)^3 = 3k^3+9k^2+15k+9$

4. Show that the equation is true for $n = k+1$

   $(k+1)^3+(k+2)^3+(k+3)^3=3k^3+18k^3+42k+36$

And this is the point where I am stuck. With a simpler formula I would simply replace the $k$ with $9M $ but how can you do it in this case?

Answer 1

Here is a solution without induction.

Let $n-1$, $n$, and $n+1$ be three consecutive numbers. As such, we have $$(n-1)^3+n^3+(n+1)^3=3n(n^2+2).$$

Divisibility by $9$ requires divisibility by $3$, twice. Hence, it suffices to show that either $n$ or $n^2+2$ is a multiple of $3$.

For $n\equiv0\pmod3$, we have $$n=3k.$$ For $n\equiv\pm1\pmod3$, we have $$n^2+2=(3k\pm1)^2+2=9k^2\pm6k+3=3(2k^2\pm2k+1).$$

Answer 2

If you want a proof with induction:

Your proof is almost correct, you're just missing the final step:

$3k^3+18k^2+42k+36 = (3k^2 + 9k^2 +15k + 9) + (9k^2 +27k + 27)$.

Note that the former is divisible by 9 by our induction hypothesis, and the latter can be written as $9(k^2 + 3k + 3)$, which is also divisible by 9.

Since the sum of two numbers divisible by 9, is also divisible by 9, we have that

$3k^3 + 18k^2 + 42k+36$ is also divisible by 9.

Answer 3

For all $n \in \{1,2,3,\dots\}$, $n^3 + (n+1)^3 + (n+2)^3$ is a multiple of $9$.

• It's true for $n=1$.

$\qquad 1^3 + 2^3 + 3^3 = 1 + 8 + 27 = 36$, which is a multiple of $9$.

• If its true for $n=k$, then it's true for $n=k+1$.

Suppose $k^3 + (k+1)^3 + (k+2)^3 = 9x$. Then

\begin{align} (k+1)^3 + (k+2)^3 + (k+3)^3 &= [k^3 + (k+1)^3 + (k+2)^3] + (k+3)^3 - k^3 \\ &= 9x + (k+3)^3 - k^3 \\ &= 9x + (k^3 + 9k^2 + 27k + 27) - k^3 \\ &= 9x + 9k^2 + 27k + 27 \\ &= 9(x + k^2 + 3k + 3) \end{align}