Proving the Largest Natural Number m for Divisibility of n^3-n

[hunt_mat]

You can easily show that $n^{3}-n$ is even for all natural numbers $n$, ehast is the highest divisor of an even number?

 

[tiny-tim]

… ehast is the highest divisor of an even number?

i know west is west, but ehast is ehast ?

 

[hunt_mat]

It should have said, what is the highest divisor of an even number

 

[Punkyc7]

so 2 is the highest divisor of n^3-n?

 

[micromass]

so 2 is the highest divisor of n^3-n?

Is 2 the highest divisor of 24?

 

[hunt_mat]

let n>4 be an even number, then it can be written n=2k, what is the highest divisor of n?

 

[Punkyc7]

no 12 was

hunt_mass would it be k?

 

[hunt_mat]

It would be. You are almost at your proof.

 

[Punkyc7]

so its (n^3-n)/2=k


now do you do induction on k or n?
im thinking n

 

[hunt_mat]

basically, once you show it's even (you have not yet done this) then that would be your answer.

 

[Punkyc7]

do i have to show n^3-n is even or (n+1)^3-(n+1) is even or both?

 

[hunt_mat]

show that it is even for arbitrary n, so I would write n=2k+1 for example...

 

[Punkyc7]

but which one do i show is even the induction one or the original one?

 

[hunt_mat]

They are both the same. Work with $n^{3}-n$ for now.

 

[Punkyc7]

so (2k)^3-2k= 2k((2k)^2-1) which is even


(2k+1)^3-(2k+1)=(2k+1)((2k+1)^2-1)

(2k+1) is always odd

((2k+1)^2-1)=4k^2+4k+1-1=2k(2k+2) which is even

since an even times an odd is even
this implies n^3-n is always even

 

[I like Serena]

I've been studying the problem statement and I agree with tiny-tim (^TM).

The only way the problem makes some "sense" to me, is if you try to find the greatest number m which divides n³-n for all n.

 

[hunt_mat]

As this has been shown to be an even number, the highest value must be half that amount.

 

[Punkyc7]

wait... so is it half of the amount or 2 right...

 

[hunt_mat]

depending on which is the bigger value.

 

[Alex B]

Sorry to dig up an old thread, but I just saw this interesting problem in Bartle & Sherbert's textbook and did a Google search to see if anyone got the same answer as me.

I'm not sure if the problem was ever stated clearly, but it is: find the largest natural number m such that m divides n^3 - n FOR ALL natural numbers n. Notice that n is not a free variable; it is bounded to a universal quantifier. Therefore, it does not make sense to have an answer that depends on n, which is undefined outside of that quantified statement. In other words, our answer should be an EXPLICIT natural number.

Now, suppose we have our answer m. So no matter what n we choose, m divides n^3 - n. For example, let's choose n = 2. Then n^3 - n = 2^3 - 2 = 6. Thus, m divides 6. But the only positive divisors of 6 are 1, 2, 3, and 6.

Note that 1 divides every number, so m obviously exists, and also m >= 1. What is left is to figure out which (if any) of the numbers 2, 3, or 6 has the property that it is a divisor of n^3 - n FOR ALL natural numbers n (using induction). Finally, our answer m is the largest of the numbers that have that property.

 

[frobeniustaco]

Digging up something old, but I too was looking through Bartle & Sherbert's Intro to Real Analysis and thought this was interesting so I googled the problem. I am with Alex B on this except for one thing: Technically, for the proof to be correct, one would have to take note of the n=1 case as well. This does turn out to be trivial, in that 1^3 - 1 = 0 is divisible by any number. It's still an important fact to not for correctness though, since we'd be interested in proving this for all Natural numbers.