Proving n = m^3 - m is a Multiple of 6: A Simple Guide

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In summary: I don't understand what your suggestion is trying to make me do, and I don't see why that's divisible by 6 (though I tried a few examples).ok I understand why that divides by 6But that is much more complicated than a direct proof so I cannot imagine why you would not use the simpler direct proof.
  • #1
phospho
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is n = m^3 - m for some integer m, then n is a multiple of 6

how do I proof this?

I tried to do it by condradiction by assuming n is not a multiple of 6, i.e. when n is divided it is in the form of n = 1+k, or n = 2+k or n = 3+k or n = 4+k or n=5+k for some integer k

however I wasn't able to get anywhere, any help?
 
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  • #2
By 1+k or 2+k or 3+k I think you mean 1+6k or 2+6k etc.

You might want to try doing this with m instead of n.
 
  • #3
Office_Shredder said:
By 1+k or 2+k or 3+k I think you mean 1+6k or 2+6k etc.

You might want to try doing this with m instead of n.

I don't get it, why?
 
  • #4
Factor m^3- m. What do you get?
 
  • #5
HallsofIvy said:
Factor m^3- m. What do you get?

m(m^2-1) = m(m-1)(m+1)
 
  • #6
phospho said:
I don't get it, why?

Did you try doing it? What did you get?

Factoring is actually a much cleaner way of doing this than my suggestion... can you explain why (m-1)(m)(m+1) is divisible by 6?
 
  • #7
Office_Shredder said:
Did you try doing it? What did you get?

Factoring is actually a much cleaner way of doing this than my suggestion... can you explain why (m-1)(m)(m+1) is divisible by 6?

I don't understand what your suggestion is trying to make me do, and I don't see why that's divisible by 6 (though I tried a few examples).
 
  • #8
ok I understand why that divides by 6

but I think the book wants me to use contradiction - how would I do that?
 
  • #9
Okay, you understand that [itex]n= m^3- m= (m-1)(m)(m+1)[/itex], the product of three consecutive numbers. So at least one of those (in fact, one of any two consecutive numbers) is even and one of them is a multiple of 3. That proves that n is divisible by 6 and I don't see why you would want any proof.

I suppose you could say "Suppose n is not divisible by 6. Then either it is not divisible by 2 or it is not divisible by 3.

Since 2 is a prime number, if n is not divisible by 2, no factor of n is not divisible by 2. In particular, m- 1 is not divisible by 2. In that case m- 1 is odd: m-1= 2k+ 1 for some integer k. Then m= m-1+ 1= 2k+ 1+ 1= 2k+ 2= 2(k+ 1) is divisible by 2, a contradiction.

Since 3 is a prime number, if n is not divisible by 3, no factor of n is divisible by 3. In particular, m-1 is not divisible by 3 which means m-1= 3k-1 or m-1= 3k+1 for some integer k. If m-1= 3k- 1 then m= m-1+1= 3k-1+ 1= 3k is divisible by 3, a contradiction. If m-1= 3k+ 1, then m+1= m-1+ 2= 3k+1+ 2= 3k+ 3= 3(k+1) is divisible by 3, a contradiction."

But that is much more complicated than a direct proof so I cannot imagine why you would not use the simpler direct proof.
 

FAQ: Proving n = m^3 - m is a Multiple of 6: A Simple Guide

How can I prove that n = m^3 - m is a multiple of 6?

To prove that n = m^3 - m is a multiple of 6, we need to show that it is divisible by 6 without leaving a remainder. This can be done by using mathematical induction, where we first show that the formula is true for m=1, and then use the assumption that it is true for m=k to prove that it is also true for m=k+1.

What is the significance of proving that n = m^3 - m is a multiple of 6?

Proving that n = m^3 - m is a multiple of 6 has practical applications in fields such as cryptography, where understanding the divisibility of numbers is important for creating secure codes. It also helps in understanding and solving other mathematical problems that involve multiples and divisibility.

Can this proof be generalized to other values besides 6?

Yes, this proof can be generalized to other values. In fact, the same method can be used to prove that n = m^3 - m is a multiple of any positive integer. The key is to use mathematical induction and show that the formula holds true for the given value of m and then use that assumption to prove it for the next value of m.

Are there any real-life examples that can be explained using this proof?

Yes, there are many real-life examples that can be explained using this proof. For instance, if you have 6 identical objects and you want to distribute them equally among a group of people, you can use this proof to show that the number of objects can be divided equally among all the people without leaving any leftovers.

Is there an easier way to prove that n = m^3 - m is a multiple of 6?

There are other methods that can be used to prove that n = m^3 - m is a multiple of 6, such as using modular arithmetic or divisibility rules. However, using mathematical induction is the simplest and most straightforward way to prove this formula. It may seem complicated at first, but once you understand the concept, it becomes an easy and effective method for proving divisibility.

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