Is my inductive proof correct?, $(n-2)^3 + (n-1)^3 + n^3 = 9 \, (x)$

Question

The proof is the sum of the cubes of three consecutive natural numbers that are divisible in 9 so $(n-2)^3 + (n-1)^3 + n^3 = 9 \, (x)$ for some $x \in \mathbb{N}$

1. Base Case: We need to show that the statement is true for the smallest possible value, which in this case is $ n = 2 $. Substituting $ n = 2 $ into the equation, we get:

   $ (2-2)^{3}+(2-1)^{3}+2^{3} = 0^{3}+1^{3}+2^{3} = 0+1+8 = 9 $

   This is divisible by 9, so the base case is true.

2. Inductive Step: We assume that the statement is true for some arbitrary natural number $ k $, i.e.,

   $ (k-2)^{3}+(k-1)^{3}+k^{3} = 9y $ for some $ y \in \mathbb{N} $.

   We need to show that the statement is also true for $ k+1 $, i.e.,

   $ ((k+1)-2)^{3}+((k+1)-1)^{3}+(k+1)^{3} = 9z $ for some $ z \in \mathbb{N} $.

   Simplifying the left-hand side, we get:

   $ (k-1)^{3}+k^{3}+(k+1)^{3} $

   We can rewrite this as:

   $ (k-2)^{3}+(k-1)^{3}+k^{3} + 3k^{2} + 3k + 1 $

   By the inductive hypothesis, we know that $ (k-2)^{3}+(k-1)^{3}+k^{3} = 9y $, so we can substitute this in:

   $ 9y + 3k^{2} + 3k + 1 $

   Factoring out a 9, we get:

   $ 9(y + k^{2} + k + 1/9) $

   Since $ y, k^{2}, k, $ and $ 1/9 $ are all natural numbers, their sum is also a natural number. Let's call it $ z $. So we have:

   $ 9z $

   This shows that the statement is also true for $ k+1 $ if it is true for $ k $.

Therefore, by mathematical induction, the sum of the cubes of any three consecutive natural numbers is divisible by 9.

Answer 1

It is the right idea, but something went wrong in the inductive step. I think it is just and algebra error.

Base case is good.

Here is how I would do the inductive step.

Assume $9|(k-2)^3+ (k-1)^3 + k^3$

Show that $9|(k-1)^3+ k^3 + (k+1)^3$ using the assumption above.

$(k-1)^3+ k^3 + (k+1)^3\\ (k-1)^3+ k^3 + ((k-2)+3)^3\\ (k-1)^3+ k^3 + (k-2)^3 + 9(k-2)^2 + 27(k-2) + 27$

From the inductive hypothesis, we are assuming that $9$ divides $(k-1)^3+ k^3 + (k-2)^3.$ And, each of the remaining terms has a factor that is divisible by 9.

I see your mistake....

You are fine at this line
$(k−1)^3+k^3+(k+1)^3$

In the next line you expand $(k-1)^3$ but you change the $(k−1)^3+k^3$ to $(k-2)^3+(k-1)^3$ without justification.

Answer 2

Alternative approach:

$~(n+3)^3 - n^3 = 9n^2 + 27n + 27 \implies 9 ~| ~[~(n+3)^3 - n^3].$

Therefore, given any $~3~$ non-negative integers, $~a,b,c~$ if $~9~$ divides $~a^3 + b^3 + c^3,~$ then (for example) $~9~$ must divide $~(a+3)^3 + b^3 + c^3.~$

So, you have that $~0^3 + 1^3 + 2^3 = 9.~$

Then, by induction, suppose that $~9~$ divides $~n^3 + (n+1)^3 + (n+2)^3.~$

Then, from the previous analysis, $~9~$ must divide

$(n+3)^3 + (n+1)^3 + (n+2)^3.$