# Primes

• Sep 7th 2012, 11:53 AM
brucewayne
Primes
The numbers $\displaystyle 9^n\pm\2$ are both primes for n=1 and n=2 in which the pairs are 7, 11 and 79, 83. Prove $\displaystyle 9^n-2$ is composite for infinitely many positive integers n and prove $\displaystyle 9^n+2$ is composite for infinitely many n.

Note: This my first LaTex post, and I was pretty excited even though it is a simple equation.
• Sep 7th 2012, 01:41 PM
MaxJasper
Re: Primes
Quote:

Originally Posted by brucewayne
The numbers $\displaystyle 9^n\pm\2$ are both primes for n=1 and n=2 in which the pairs are 7, 11 and 79, 83. Prove $\displaystyle 9^n-2$ is composite for infinitely many positive integers n and prove $\displaystyle 9^n+2$ is composite for infinitely many n.

Values of n<1000 that make at least one of these two $\displaystyle 9^n\pm 2 = \text{Prime}$

n ={1, 2, 3, 4, 5, 7, 11, 12, 13, 18, 45, 49, 51, 55, 63, 193, 247, 260, 324, 390, 393, 610, 929,....}
• Sep 7th 2012, 01:45 PM
topsquark
Re: Primes
Quote:

Originally Posted by brucewayne
The numbers $\displaystyle 9^n\pm\2$ are both primes for n=1 and n=2 in which the pairs are 7, 11 and 79, 83. Prove $\displaystyle 9^n-2$ is composite for infinitely many positive integers n and prove $\displaystyle 9^n+2$ is composite for infinitely many n.

Note: This my first LaTex post, and I was pretty excited even though it is a simple equation.

Consider 9^n mod 7... (There might be a better mod, but this one works, anyway.)

-Dan
• Sep 7th 2012, 06:53 PM
richard1234
Re: Primes
Modulo 7,

$\displaystyle 9^1 \equiv 2$

$\displaystyle 9^2 \equiv 4$

$\displaystyle 9^3 \equiv 1$

...and it repeats. Here, you can prove that there are infinitely many n such that $\displaystyle 9^n - 2 \equiv 0 (\mod 7)$, but you can't say anything about $\displaystyle 9^n + 2$.

But if we look at it mod 11, it turns out that $\displaystyle 9^1$, $\displaystyle 9^6$, $\displaystyle 9^{11}$, etc. are 9 mod 11, adding 2 makes them a multiple of 11.