# disproving prime numbers

• September 27th 2008, 02:22 AM
andrewuk
disproving prime numbers
how do you prove that number is not Prime?

the example in my book says prove that 2 to the power of 66 is not prime and then says the keys lies in this simple algebra $x squared-1=(x-1)(x+1)$ it says to use that to write a proof that the number is not prime so what am supposed to do with that equation?
• September 27th 2008, 02:46 AM
Moo
Hello,
Quote:

Originally Posted by andrewuk
how do you prove that number is not Prime?

the example in my book says prove that 2 to the power of 66 is not prime and then says the keys lies in this simple algebra $x squared-1=(x-1)(x+1)$ it says to use that to write a proof that the number is not prime so what am supposed to do with that equation?

huh
:confused: 2 divides $2^{66}$, therefore it is not prime :confused:

Another way :
$2^{66}-1=(2^{33}-1)(2^{33}+1)$

$\implies 2^{66}=(2^{33}-1)(2^{33}+1)+1$

But since $2^{33}$ is an even number, $2^{33}-1 \text{ and } 2^{33}+1$ are odd numbers. Thus their product is an odd number.
If we substract 1 to it, we get an even number. Hence $2^{66}$ is an even number (and different from 2), so it is not prime.

But that's finding the most complicated way...
• September 27th 2008, 04:22 AM
andrewuk
Ok thanks for that, don't know why the book couldn't of included an example first to show how the equation is used.

How would i split it if the power is an odd number? e.g. 2 to the power of 67
• September 27th 2008, 06:12 AM
CaptainBlack
Quote:

Originally Posted by Moo
Hello,

huh
:confused: 2 divides $2^{66}$, therefore it is not prime :confused:

Another way :
$2^{66}-1=(2^{33}-1)(2^{33}+1)$

$\implies 2^{66}=(2^{33}-1)(2^{33}+1)+1$

But since $2^{33}$ is an even number, $2^{33}-1 \text{ and } 2^{33}+1$ are odd numbers. Thus their product is an odd number.
If we substract 1 to it, we get an even number. Hence $2^{66}$ is an even number (and different from 2), so it is not prime.

But that's finding the most complicated way...

This is ridiculus, the very definition of $2^{66}$ makes it composite and hence not prime as it has at least three distinct divisors. So any proof that $2^{66}$ is not prime that uses any of the laws of powers is using this, and so is just a complicated version of the simple proof.

RonL
• September 27th 2008, 07:59 AM
andrewuk
Ye the example does seem stupid but i still need to know how to use that equation when the power is an odd number, can you have a decimal power?
• September 27th 2008, 08:47 AM
bobak
Its considered rude to post the same question on two forums.

Prime numbers - The Student Room

Bobak
• September 28th 2008, 11:00 AM
hcir614
Given any number A with prime factorization $x_0^{n_0} \cdot x_1^{n_1} \cdot \dots \cdot x_k^{n_k}$ is composite with $(n_0 + 1) \cdot (n_1 + 1) \cdot \dots \cdot (n_k + 1)$ factors.

So, this works for any prime factorization, whether it is to an odd power or not.

I do not quite understand what you mean by decimal power. Do you mean can you have $2^{0.3}$? If so, that number does exist, however you will not find factors for it.
• September 29th 2008, 09:34 AM
wisterville
Hello,

Here is my opinion: someone made an error when copying.
The real question in the book should be "Prove that $2^{66}-1$ is a prime."

Bye.
• September 29th 2008, 01:43 PM
CaptainBlack
Quote:

Originally Posted by wisterville
Hello,

Here is my opinion: someone made an error when copying.
The real question in the book should be "Prove that $2^{66}-1$ is a prime."

Bye.

I'm sorry now I'm lost.

$2^{66}-1$ is not prime as:

$2^{66}-1=(2^{33}+1)(2^{33}-1)$

RonL
• October 1st 2008, 05:22 AM
wisterville
Hello,

Sorry, I wanted to say:
The real question in the book should be "Prove that http://www.mathhelpforum.com/math-he...e1815463-1.gif is NOT a prime."

Bye.