# Proof involving prime numbers

• Mar 5th 2011, 09:09 PM
demode
Proof involving prime numbers
I need to prove that if either of the numbers $\displaystyle 2^n-1$ and $\displaystyle 2^n+1$ is prime for $\displaystyle n>2$, then the other one is not.

Attempt: If we name them $\displaystyle a=2^n-1$ and $\displaystyle b=2^n+1$, we realize that $\displaystyle b=a+2$. And the the smallet prime for the first number is when n=3, so $\displaystyle a=2^3-1=7$. For the other number the first prime occurs at n=4 and it's 17.

So, how should I prove this question? If we suppose $\displaystyle 2^n-1$ is prime, it will have only the two divisors $\displaystyle \{ 1, 2^n-1 \}$. Then what does that say about $\displaystyle 2^n+1$?

(We can't say that the second number is not prime because it is two digits larger than the first one, since many prime numbers are distrubuted at intervals of 2. And there is a chance it can be prime since it will be odd).
• Mar 5th 2011, 10:41 PM
Chris11
Look mod 3.
• Mar 7th 2011, 12:12 AM
demode
Quote:

Originally Posted by Chris11
Look mod 3.

What do you mean "look mod 3"? Please explain.
• Mar 7th 2011, 01:14 AM
chisigma
Quote:

Originally Posted by demode
I need to prove that if either of the numbers $\displaystyle 2^n-1$ and $\displaystyle 2^n+1$ is prime for $\displaystyle n>2$, then the other one is not.

Attempt: If we name them $\displaystyle a=2^n-1$ and $\displaystyle b=2^n+1$, we realize that $\displaystyle b=a+2$. And the the smallet prime for the first number is when n=3, so $\displaystyle a=2^3-1=7$. For the other number the first prime occurs at n=4 and it's 17.

So, how should I prove this question? If we suppose $\displaystyle 2^n-1$ is prime, it will have only the two divisors $\displaystyle \{ 1, 2^n-1 \}$. Then what does that say about $\displaystyle 2^n+1$?

(We can't say that the second number is not prime because it is two digits larger than the first one, since many prime numbers are distrubuted at intervals of 2. And there is a chance it can be prime since it will be odd).

In...

Twin Primes -- from Wolfram MathWorld

... is written that, with the only exception of $\displaystyle (3,5)$, all the twin primes are of the form $\displaystyle 6\ k \pm 1$. But if $\displaystyle n>2$ the equation $\displaystyle 2^{n} = 6\ k$ has no solutions for n and k integer so that...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Mar 7th 2011, 02:38 PM
Chris11
Quote:

Originally Posted by demode
What do you mean "look mod 3"? Please explain.

Look at remainders upon division by 3