Proof That Cubes Of Three Consecutive Integers Is Divisible By 9

We need to show that the sum of the cubes \(n-1\), \(n\) and \(n+1\) is divisible by 9. We will use mathematical induction to prove this. Consider the integers -1, 0 and 1. If we sum their cubes:

$$\begin{gather} (-1)^3 + 0^3 + 1^3 = \\ -1 + 0 + 1 = 0 \end{gather}$$

This shows that the sum of the cubes of \(n-1\), \(n\) and \(n+1\) is divisible by 9 when \(n=0\).

Now consider \((n+k)^3\):

$$ (n+k)^3 = n^3 + 3kn^2 + 3nk^2 + k ^3 $$

This means:

$$\begin{align} k = -1 &\implies n^3 - 3n^2 + 3n -1 \\ k=0 &\implies n^3 \\ k = 1 &\implies n^3 + 3n^2 + 3n + 1 \end{align}$$

Summing them we get:

$$ (n-1)^3 + n^3 + (n+1)^3 = 3n^3 + 6n $$

Let's call this expression \(f(n)\):

$$ f(n) = 3n^3 + 6n $$

We have already shown that \(f(n)\) is divisible by 9 when \(n=0\). Now replace \(n\) with \(n+1\):

$$\begin{align} f(n+1) &= 3(n+1)^3 + 6(n+1) \\ &= 3(n^3+3n^2+3n+1) + 6(n+1) \\ &= 3n^3+9n^2+9n+3 + 6n+6 \\ &= (3n^3+6n)+9n^2+ 9n + 9 \\ &= f(n) + 9n^2+ 9n + 9 \end{align}$$

We know that \((9n^2+ 9n + 9)\) is divisible by 9 because all the terms have a factor of 9. This means if \(f(n)\) is divisible by 9, then so it \(f(n+1)\). Since \(f(0)\) is divisible by 9, then so is \(f(1)\). Since \(f(1)\) is divisible by 9, then so is \(f(2)\), and so on.

We have proved that \(f(n)\) is divisible by 9 for \(n \ge 0\). Now let's focus on the negative integers. Replace \(n\) with \(n-1\):

$$\begin{align} f(n-1) &= 3(n-1)^3 + 6(n-1) \\ &= 3(n^3-3n^2+3n-1) + 6(n-1) \\ &= 3n^3-9n^2+9n-3 + 6n-6 \\ &= (3n^3+6n)-9n^2+ 9n -9 \\ &= f(n) - 9n^2+ 9n -9 \end{align}$$

If \(f(n)\) is divisible by 9, then \(f(n-1)=f(n) - 9n^2+ 9n -9\) is divisible by 9 aswell. Since \(f(0)\) is divisible by 9, then so is \(f(-1)\). Since \(f(-1)\) is divisible by 9, then so in \(f(-2)\), and so on. Therefore, \(f(n)\) is divisible by 9 for all negative integers.

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