Why Is \( n^3 - n \) Always Divisible by 3?

[um0123]

Polynomials divisible by...

Homework Statement

1) Explain why

$$n^3 - n$$

is always divisible by 3 for any n that is an element of the natural numbers.

2) Give 2 other polynomials that are always divisble by 3.

3) Give 2 polynomials that are divisible by 2 but not 4, and 2 other polynomials that are divisible by 2 and 4.

Homework Equations

$$n^3 - n$$

The Attempt at a Solution

Im just really confused and don't know where to start with this one. I just need some hints to finding out why $$n^3 - n$$ is divislble by three. And some help for finding a method for finding polynomials divisible by a specific number (if such a method exists).

 

[Dick]

In this case n^3-n=(n-1)*n*(n+1). What does that tell you? Can you generalize that to find a polynomial divisible by any number?

 

[um0123]

the factored form $$n(n-1)(n+1)$$ tells me that there are zeros at 1 and -1 and 0. But i don't understand what you mean by "generalize that to find a polynomial divisible by any number."

 

[nietzsche]

what happens when you multiply any three consecutive integers? what properties will the product have?

 

[Dick]

the factored form $$n(n-1)(n+1)$$ tells me that there are zeros at 1 and -1 and 0. But i don't understand what you mean by "generalize that to find a polynomial divisible by any number."

We aren't talking about where the zeros are, we are talking about divisibility by 3. Another way to check is to evaluate n^3-n mod 3 for n=0, 1 and 2. Have you done stuff like that?

 

[um0123]

We aren't talking about where the zeros are, we are talking about divisibility by 3. Another way to check is to evaluate n^3-n mod 3 for n=0, 1 and 2. Have you done stuff like that?

we haven't learned modulo in class, but i know what it is from my experience with c++. I don't know if he expects us to use it, but as long as i solve the problem in a manner that he sees how i got my answer it is fine if i do that.

if i check the mod of 3 for 0,1, and 2, it doesn't prove for when n goes to infinity though.

 

[slider142]

we haven't learned modulo in class, but i know what it is from my experience with c++. I don't know if he expects us to use it, but as long as i solve the problem in a manner that he sees how i got my answer it is fine if i do that.

if i check the mod of 3 for 0,1, and 2, it doesn't prove for when n goes to infinity though.

You don't really need the modulus argument. Can you find three consecutive integers that are not divisible by 3? If so, what are they? If not, why not? As a hint, note that 3 consecutive integers can always be written in the form n, n+1 and n+2 for some n. Can you show what happens if none of them are divisible by 3? (This is a lead to a proof by contradiction argument)

 

[um0123]

i don't understand why n, n+1, and n+2 is involved. Is that the same as n, n-1, n+1? but how do i logically prove that its always divisible by three up to inifinity, and how am i supposed to come up with my own polynomials?

 

[nietzsche]

if you have three consecutive integers, can you always factor out a three? yes, because if there are three consecutive integers, one of them has to be a multiple of 3, because you're choosing three consecutive integers.

 

[um0123]

if you have three consecutive integers, can you always factor out a three? yes, because if there are three consecutive integers, one of them has to be a multiple of 3, because you're choosing three consecutive integers.

but you can't factor out a 3 from all of them, only the one divisible by three, so i am confused on how the entire thing is able to factor out a three.

 

[slider142]

but you can't factor out a 3 from all of them, only the one divisible by three, so i am confused on how the entire thing is able to factor out a three.

?? If b is divisible by 5, then abc is also divisible by 5 irrespective of whether a or c are divisible by 5.

 

[um0123]

?? If b is divisible by 5, then abc is also divisible by 5 irrespective of whether a or c are divisible by 5.

i don't doubt you, but i don't see how that works.

 

[nietzsche]

well...take for example 4, 5, 6. you can write 6 as 3x2.

so 4x5x6 = 4x5x3x2 and if you divide that by 3, you get 4x5x2.

you can write any number that is divisible by 3 as 3k, where k is an integer.

 

[um0123]

OH, i get it now. I was just being really dumb. okay, so to make polynomials divisible by a specific number all i need to do is have part of it divisible by that number.

i want to try this now, but i got to go...:(

 

[slider142]

i don't doubt you, but i don't see how that works.

Straightforward. If b is divisible by 5, then b = 5*n for some n. Likewise, abc = a*5*n*c = 5*a*n*c so abc is divisible by 5.

 

[um0123]

im having trouble coming up with polynomials that are divisible by 2 but not 4.

 

[Bohrok]

What are the terms in the sequence of numbers divisible by 2 but not 4?

 

[um0123]

2, 6, 10, 14...

but i tried to put them in and it didnt work out.

 

[Dick]

2, 6, 10, 14...

but i tried to put them in and it didnt work out.

Put them into what? Can't you find a polynomial p(x) such that p(1)=2, p(2)=6, p(3)=10 etc?

 

[Bohrok]

2, 6, 10, 14...

but i tried to put them in and it didnt work out.

Try graphing those points and find a linear equation that fits those points.

 

[um0123]

got it, thanks for the help everyone!

 

[Mark44]

An integer that is divisible by 2 but not by 4 has only a single factor of 2 in its factorization. An integer that is divisible by 4 has at least two factors of 2 in it factorization. Can you come up with a polynomial that has only one factor of 2?