Proving Divisibility of n^3-n by 6

In summary, the conversation is discussing how to prove that n^3 - n is divisible by 6 for nonnegative integers. The suggestion is to use mathematical induction, but the speaker suggests taking a simpler approach by factoring n^3 - n. This will immediately show that 6 is always a factor. The speaker also mentions struggling with algebraic skills and asks for clarification on their approach.
  • #1
Goldenwind
146
0

Homework Statement


Prove that n^3 - n is divisible by 6, when n is a nonnegative integer.


The Attempt at a Solution


Mathematical induction:

It works for n=0
It works for n=1 (Extra step, just in case)
Check if it works for the (k+1)th step.

For it to work, it must be expressible as 6x, where x is some integer.

In other words, to prove: (k+1)^3 - k = 6x

Can someone nudge me on this? I'm either making a mistake by calling it 6x, and maybe it should be 6k or something else...

...and/or, I'm just missing the algebraic skills to change LS into RS.
 
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  • #2
Take the lazy way!

Goldenwind said:
Prove that n^3 - n is divisible by 6, when n is a nonnegative integer.

The attempt at a solution
Mathematical induction

No no no no no!

Far too amibitious!

Take the lazy way!

Just factorise [tex]n^3 - n[/tex], and you'll immediately see why 6 is always a factor! :smile:

Ping!
 

1. How do you prove the divisibility of n^3-n by 6?

To prove the divisibility of n^3-n by 6, we can use the fact that any integer can be written as 6k, 6k+1, 6k+2, 6k+3, 6k+4, or 6k+5 for some integer k. Since n^3-n is always even, we can rewrite it as (6k+1)^3-(6k+1) = 6(6k^2+6k). This shows that n^3-n is indeed divisible by 6.

2. Can you use induction to prove the divisibility of n^3-n by 6?

Yes, we can use induction to prove the divisibility of n^3-n by 6. First, we can verify that the statement is true for n=1. Then, assuming it is true for some arbitrary integer k, we can show that it is also true for k+1. This proves that the statement holds for all positive integers, and therefore n^3-n is divisible by 6 for all n.

3. What is the significance of proving the divisibility of n^3-n by 6?

Proving the divisibility of n^3-n by 6 is significant because it is a useful mathematical tool in various fields such as number theory and cryptography. It also helps us understand the relationship between divisibility and patterns in numbers.

4. Can the divisibility of n^3-n by 6 be extended to other values besides 6?

Yes, the divisibility of n^3-n can be extended to other values besides 6. In fact, any integer k can be used in place of 6, as long as k is a factor of both n and n^3. This is because if k is a factor of n, then n^3 is also divisible by k, and thus n^3-n is divisible by k.

5. How can the divisibility of n^3-n by 6 be applied in real-world situations?

The divisibility of n^3-n by 6 can be applied in various real-world situations, such as in coding and encryption algorithms. It can also be used in number theory to study the properties of integers and identify patterns in numbers. Additionally, it has applications in fields such as computer science and engineering.

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