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Thread: Primes

  1. #1
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    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.
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    Senior Member MaxJasper's Avatar
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    Lightbulb Re: Primes

    Quote Originally Posted by brucewayne View Post
    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,....}
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  3. #3
    Forum Admin topsquark's Avatar
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    Re: Primes

    Quote Originally Posted by brucewayne View Post
    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
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    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.
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