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Answer: Cubes of three Consecutive Integers

Saturday, 9 of February , 2008 at 7:05 pm

Question: Show that the sum of the cubes of any three consecutive integers is divisible by 9.

Answer: In order to solve this problem, I am going to define the questions a little more formally.

Prove that there \exists a k \in \mathbb Z where f(n) = 9k given

f(n) = (n)^3 + (n+1)^3 + (n+2)^3 and n \in \mathbb Z

I will be solving this problem using induction.

Base Case:
First we need show that the problem is true for the base case (n=0)

f(0) = 0^3 + (0 + 1)^3 + (0 + 2)^3 = 0 + 1 + 8 = 9

Therefore, there \exists a k \in \mathbb Z (which would be k = 1) where f(0) = 9k.

Induction Step (for positive values):
In order to prove that the problem holds for n \in \mathbb N, we need to show:

If f(n) = 9k ~~ \Longrightarrow ~~ \exists~a~j \in \mathbb Z where f(n+1) = 9j where n,k \in \mathbb Z

In order to prove this statement, we need to reduce f(n+1).

f(n+1) = (n+1)^3 + (n+2)^3 + (n+3)^3
= n^3 + (n+1)^3 + (n+2)^3 + (n+3)^3 - n^3
= f(n) + (n+3)^3 - n^3
= f(n) + (n+3)(n+3)(n+3) - n^3
= f(n) + (n^3 + 9n^2 +27n + 27) - n^3
= f(n) + 9(n^2 +3n + 3)
= 9k + 9(n^2 +3n + 3)
= 9(n^2 +3n + 3 + k)

Since (n^2 +3n + 3 + k) \in \mathbb Z, we have shown if f(n) = 9k it implies that \exists~a~j \in \mathbb Z where f(n+1) = 9j.

Induction Step (for negative values):
Since we already proved by induction that the problem holds for all natural numbers, all we need to do to extend this proof to all integers is to show that:

If f(n) = 9k ~~ \Longrightarrow ~~ \exists~a~j \in \mathbb Z where f(n-1) = 9j where n,k \in \mathbb Z

In order to prove this statement, we need to reduce f(n-1).

f(n-1) = (n-1)^3 + n^3 + (n+1)^3
= (n-1)^3 - (n+2)^3 + n^3 + (n+1)^3 + (n+2)^3
= f(n) + (n-1)^3 - (n+2)^3
= f(n) + (n^3 - 3n^2 + 3n - 1) + (n^3 + 6n^2 + 12n + 8)
= f(n) + (-9n^2 - 9n - 9)
= f(n) + 9(-n^2 - n - 1)
= 9k + 9(-n^2 - n - 1)
= 9(-n^2 - n - 1 + k)

Since (-n^2 - n - 1 + k) \in \mathbb Z, we have shown if f(n) = 9k it implies that \exists~a~j \in \mathbb Z where f(n-1) = 9j.

Therefore, we have proven by induction that the sum of the cubes of any three consecutive integers is divisible by 9.

Category: Algebra, Answers

2 Comments

Comment by Vishal

Made Friday, 22 of February , 2008 at 8:11 am

Hi

There is a small typo in the problem statement. There should be a (n+2)^3 instead of (n+3)^3.

A different approach - and I am sure you are familiar with it - would be to use modular arithmetic. This has the advantage that we don’t need to analyze separate cases for positive and negative integers. The disadvantage, of course, is that there are lots of if…then cases albeit short ones.

Comment by eldila

Made Friday, 22 of February , 2008 at 9:53 am

Thanks for catching the typo. I started solving the problem using modular arithmetic, but it started getting uglier than I thought the problem should be. Although you need to handle two cases for my approach, the actually math behind the problem is trivial.
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