Proving Divisibility Property for Odd Integers: 24 l a(a2-1)

[annoymage]

Homework Statement

prove the following

if a is an odd integer, then, 24 l a(a²-1)

(i'm not familiar with modulo yet, i think it can help, but let don't use it yet ;P)

Homework Equations

n/a

The Attempt at a Solution

i stumbled when using 2n+1=a for all integer n, because i will only get (2n)(4n²+4n), i can't see how it can divided by 24, should i think of any other equivalent classes of odd integer?? or any other clue? T_T help

 

[Tedjn]

For 24 to divide a(a²-1), convince yourself that this happens exactly when 8 and 3 divide evenly.

 

[annoymage]

i don't understand T_T, do you mean i need to find 8 l "something" , 3 l "something"

like 8 l a and 3 l a²-1 then 24 l a(a²-1), something like that??

 

[Tedjn]

I meant 24 | a(a²-1) if and only if 8 | a(a²-1) and 3 | a(a²-1).

 

[annoymage]

hmm, if that's the case i can proof 3 l a(a²-1)
, 4 l a(a²-1) , and 2 l a(a²-1)
will imply 8 l a(a²-1) and 3 l a(a²-1), then 24 l a(a²-1) for all odd integer a

but now this troubled me how to proof a l d and c l d then ac l d , wait let me think, don't go offline, hoho

 

[Tedjn]

Well, proving for 4 and 2 doesn't imply 8. Why not?

 

[annoymage]

i'm sorry, I'm not really good with english

before i continue, "Well, proving for 4 and 2 doesn't imply 8. Why not?" I'm not sure if this a question, or you implying that 4 and 2 does imply 8,

if you're implying, i follow your logic, then 4 and 2 does imply 8,

but now i want to verify is all a,b,c,d in Z follow this logic

if a l d and c l d then ac l d

i cannot proof this T_T, should i proof by contradiction maybe, wait, i'll try ;P

 

[annoymage]

wait I'm confused with myself

 

[Tedjn]

No, I mean 4 and 2 actually don't imply 8. Can you think of the reason why not?

 

[annoymage]

ugh, i can't find counter example (should i use proof by contradiction?),

and you said 8 and 3 does imply 24, should i use (8 and 3 and (not 24)), is a contradiction??

 

[Dick]

You really haven't explained why you think anything divides a(a^2-1). Why do you think 3 divides it? Why do you think 4 divides it? Let's start this over again.

 

[annoymage]

lets let's start again,

i mentioned

4 l a(a^2-1) for all odd integer

proof,

a=2n+1 for all integer n , then,

(2n+1)((2n+1)^2-1 = 8n^3 + 12n^2 + 4 n = 4(2n^3+3n^2+1), and it can divides by 4,

hmm, and i don't remember how i got 3 l a(a^2-1),

 

[Dick]

Here's a hint. (a^2-1)=(a+1)(a-1). So a(a^2-1)=(a-1)a(a+1). Now rethink this. You don't even need a lot of technical stuff to answer this.

 

[annoymage]

yea, just now i got it like this,

when a=3n , 3(n(3n+1)(3n-1))

when a=3n-1 , 3(n(3n-1)(3n-2))

when a=3n+1 , 3(n(3n+1)(3n+2)) for all integer n

so 3 l a(a^2-1), for all integer a, hence, 3 l a(a^2-1) for all odd integer,

is this alright?

 

[annoymage]

hmm , i just noticed that

4 l a(a^2-1) for all integer,

anyway, what next thing to do?

 

[Dick]

yea, just now i got it like this,

when a=3n , 3(n(3n+1)(3n-1))

when a=3n-1 , 3(n(3n-1)(3n-2))

when a=3n+1 , 3(n(3n+1)(3n+2)) for all integer n

so 3 l a(a^2-1), for all integer a, hence, 3 l a(a^2-1) for all odd integer,

is this alright?

I think it's ok. But maybe you are making this too hard. a-1,a,a+1 are three consecutive integers. Of course one is divisible by three. Now why is the product divisible by 8 if a is odd. Just say it in words.

 

[annoymage]

Now why is the product divisible by 8 if a is odd. Just say it in words.

more clue please ;P

 

[annoymage]

hmm, when a is odd,

even*odd*even is even?

 

[Dick]

Ok, but you shouldn't need this. If a is odd then (a-1) and (a+1) are two consecutive even numbers. Might one be divisible by 4?

 

[annoymage]

i can see that it divisible by 4, but if you asking "why", i don't know how to answer T_T

 

[Dick]

i can see that it divisible by 4, but if you asking "why", i don't know how to answer T_T

I'm asking whether you can show one of a+1 and a-1 is itself divisible by 4. Not whether the product is divisible by 4. "Two consecutive even numbers" think about it.

 

[annoymage]

owh, hmm, i only can answer like this,

a+1 divisible by 4 only when a=4n-1 , for odd integer n

a-1 divisible by 4 only when a=4n+1, for odd integer n

T_T, i only running around in circle, perhaps more clue (if you're not offended by my stupidity)

 

[Dick]

owh, hmm, i only can answer like this,

a+1 divisible by 4 only when a=4n-1 , for odd integer n

a-1 divisible by 4 only when a=4n+1, for odd integer n

T_T, i only running around in circle, perhaps more clue (if you're not offended by my stupidity)

Can't any odd number be written as either 4n-1 or 4n+1?

 

[annoymage]

Can't any odd number be written as either 4n-1 or 4n+1?

my answer is no, because every odd number can be written either 4n-1 or 4n+1

hmm, then a+1,a-1 are conservative even integer, product of two conservative integer will divisible by 4, then (a+1)(a-1) must divisible by 4? is that argument correct and sufficient?

 

[Dick]

my answer is no, because every odd number can be written either 4n-1 or 4n+1

hmm, then a+1,a-1 are conservative even integer, product of two conservative integer will divisible by 4, then (a+1)(a-1) must divisible by 4? is that argument correct and sufficient?

Yes, the product of ANY two even numbers is divisible by 4. That's correct but not sufficient. You need to show the product of two CONSECUTIVE even numbers is divisible by 8.

 

[annoymage]

i just read what consecutive means ;P
anyway

let 2n, 2n+2 be two consecutive even integer, so the product are (n)4(n+1),

let 2a, 2a+1, be consecutive integer,

if n=2a, then 8 l 8a(n+1)

if n=2a+1, then 8 l 8(a+1)n

so, 8 divides either one of them,

now back to your question (8 l a(a+1)(a-1) when a is odd)

let a=2n+1 for some integer n

then (2n+1)(2n+2)(2n), then 8 l (2n+1)(2n+2)(2n) since 2n+2, 2n are 2 consecutive even integer, so the product must divisible by 8, is that correct and sufficient?

anyway back to the question,

i still don't know why 24 divides a(a+1)(a-1)

or

" 24 | a(a2-1) if and only if 8 | a(a2-1) and 3 | a(a2-1) " how this is true if a is odd,

more clue ;P

 

[Dick]

That's a bit confusing, but I think you've got the idea. Let N be any number. You don't know why 8|N and 3|N if and only if 24|N? Think about the prime factorization of N.