## Theorem

The sum of the cubes of three consecutive natural numbers is a multiple of 9.

## Proof

First, introducing a predicate $P$ over $\mathbb{N}$, we rephrase the theorem as follows. $$\forall n \in \mathbb{N}, P(n) \quad \text{where} \quad P(n) \, := \, n^3 + (n + 1)^3 + (n + 2)^3 \text{ is a multiple of 9}$$ We prove the theorem by induction on $n$.

### Basis

Below, we show that we have $P(n)$ for $n = 0$. $$0^3 + 1^3 + 2^3 = 0 + 1 + 8 = 9 = 9 \cdot 1$$

### Inductive step

Below, we show that for all $n \in \mathbb{N}$, $P(n) \Rightarrow P(n + 1)$.

Let $k \in \mathbb{N}$. We assume that $P(k)$ holds. In the following, we use this assumption to show that $P(k + 1)$ holds.

By the assumption, there is a $i \in \mathbb{N}$ such that $i \cdot 9 = k^3 + (k + 1)^3 + (k + 2)^3$. We use this fact in the following equivalent transformation. The transformation turns the sum of cubes in the first line, for which we need to show that it is a multiple of 9, into a product of 9 and another natural number.

$(k + 1)^3 + (k + 2)^3 + (k + 3)^3 \\ = (k + 1)^3 + (k + 2)^3 + k^3 + 9k^2 + 27k + 27 \\ = k^3 + (k + 1)^3 + (k + 2)^3 + 9k^2 + 27k + 27 \quad | \text{ using the induction hypothesis} \\ = 9i + 9k^2 + 27k + 27 \\ = 9 \cdot i + 9 \cdot k^2 + 9 \cdot 3k + 9 \cdot 3 \\ = 9 \cdot (i + k^2 + 3k + 3)$

We see that the above product has precisely two factors: 9 and another natural number. Thus the product is a multiple of 9. This completes the induction.

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Looks fine to me! –  Braindead Mar 30 at 13:02
someone should make something along the lines of markExchange instead of stackExchange. Marking instead of answering sounds fresh and good to me –  Nicholas Kyriakides Mar 31 at 3:59
Looks good. You could maybe add a Halmos to the end, to be extra if not overly formal, $\square$. Check out math.stackexchange.com/questions/56606/… –  guydudebro Mar 31 at 4:42

I do not think that this is a real question but if you were my student i would give you an A.
It is all fine to me.

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It's fine, here's a simpler proof without induction:

$n^3\equiv n\ (\text{mod }3)$, because it obviously holds for $n=-1,0,1$.

Therefore $3n^3\equiv3n\ (\text{mod 9})$ and $$(n-1)^3+n^3+(n+1)^3\equiv3n^3+6n\equiv0\ (\text{mod }9)$$

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i.e. $\ {\rm mod}\ 9\!:\ 3n^3\!+6n\equiv 3(n^3\!-n)\equiv 0\$ by little Fermat. –  Bill Dubuque Mar 30 at 13:16
One might also conclude with a clarifying statement about what has been done - that the hypothesis is true for all $n \in \Bbb N$.
What do you think about the following statement? "By the basis, the inductive step, and the principle of induction; $P(n)$ is true for all $n \in \mathbb{N}$." –  draco malfoy Mar 30 at 13:51