# Math Help - disproving prime numbers

1. ## 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?

2. Hello,
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
2 divides $2^{66}$, therefore it is not prime

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...

3. 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

4. Originally Posted by Moo
Hello,

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

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

5. 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?

6. Its considered rude to post the same question on two forums.

Prime numbers - The Student Room

Bobak

7. 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.

8. 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.

9. 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

10. Hello,

Sorry, I wanted to say:
The real question in the book should be "Prove that is NOT a prime."

Bye.