# Proof help

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• Mar 3rd 2007, 10:09 AM
free_to_fly
Proof help
Let n be a positive integer:
1.Factorize n^5-n^3 (which I got to be n^2(n+1)(n-1)) and show that it's divisible by 24.
2. Prove that 2^2n is divisible by 3.
3. If n-1 is divisible by three, show that n^3-1 is divisible by nine.

I can explain using words more or less why, but I struggle to put them into mathematical language.

Thanks in advance for your help.;)
• Mar 3rd 2007, 01:16 PM
Soroban
Hello, free_to_fly!

I think I have a proof for #3 . . .

Quote:

3. If n - 1 is divisible by three, show that n³ - 1 is divisible by nine.

We are told that n - 1 is divisible by 3.
. . Hence: .n - 1 .= .3a, for some integer a.

Multiply both sides by (n² + n + 1): .(n - 1)(n² + n + 1) .= .3a(n² + n + 1)

. . and we have: .n³ - 1 .= .3a(n² + n + 1) .[1]

The quadratic factor is: .n² + n + 1 .= .(n - 1)(n + 2) + 3

. . Since n - 1 .= .3a, it becomes: .3a(n + 2) + 3 .= .3(an + 2a + 1)

. . So the quadratic factor is a multiple of 3: .3b, for some integer b.

Substitute into [1] and we have: .n³ - 1 .= .3a(3b) .= .9ab, a multiple of 9.

Therefore: .n³ - 1 is divisible by 9.

• Mar 3rd 2007, 04:50 PM
ThePerfectHacker
Quote:

Originally Posted by free_to_fly
2. Prove that 2^2n is divisible by 3.

Is there a mistake on this one?
• Mar 3rd 2007, 06:25 PM
Soroban
Hello again, free_to_fly!

Quote:

1. Factor N .= .n^5 - n^3, and show that it's divisible by 24.

We have: .N .= .n³(n - 1)(n + 1)

. . . . . . . .N .= .n²·(n-1)·n·(n+1)
. . . . . . . . . . . . . .\__________/

N contains the product of three consecutive integers.
. . The product of three consecutive integers is divisible by 3.

Now we must show that N is divisible by 8.
There are two cases to consider.

(1) n is even: .n = 2k for some integer k.
. . Then: .N .= .(2k)³(2k - 1)(2k + 1) .= .8·k³(2k - 1)(2k + 1)
Therefore, N is divisible by 8.

(2) n is odd: .n = 2k + 1 for some integer k.
. . Then: .N .= .(2k + 1)³(2k)(2k + 2) .= .4·k(k + 1)(2k + 1)³
Hence, N is divisible by 4.

Note that k and k + 1 are consecutive integers.
. . Hence, one of them is even, a multiple of 2.
Therefore, N is divisible by 8.

• Mar 4th 2007, 05:57 AM
free_to_fly
Mistakes
Yep, it was meant to be:
prove that (2^2n)-1 is divisible by three.

Thanks fo pointing that out.
• Mar 4th 2007, 08:07 AM
Soroban
Hello, free_to_fly!

Quote:

Prove that: .N .= .2^{2n} - 1 is divisible by three.

We have: .(2)^{2n} .= .(2^2)^n .= .4^n

. . Then: . N .= .4^n - 1 .= .(3 + 1)^n - 1

Binomial expansion: . N .= .(3^n + n·3^{n-1} + . . . + n·3 + 1) - 1

. . . . . . . . . . . . . . . .N .= .3^n + n·3^{n-1} + . . . + n·3

. . . . . . . . . . . . . . . .N .= .3·[3^{n-1} + n·3^{n-2} . . . + n] ... a multiple of 3

Therefore, N is divisible by 3.