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Thread: Fermat's Theorm and Divisibility

  1. #1
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    Fermat's Theorm and Divisibility

    I am trying to find n such that $\displaystyle 3^{n} + 2^{n}$ is divisible by 13.

    I am trying to let $\displaystyle n = 12x + r$ and $\displaystyle n = 12y + s$ which gives $\displaystyle 3^{12x + r} + 2^{12y + s}$. But I end up running in circles and getting nowhere!

    Any ideas?
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  2. #2
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    $\displaystyle \begin{array}{rcll} 3^1 & \equiv & 3 & (\text{mod } 13) \\ 3^2 & \equiv & 9 & (\text{mod } 13) \\ 3^3 & \equiv & 1 & (\text{mod } 13) \\ 3^4 & \equiv & 3 & (\text{mod } 13) \\ & \vdots & & \end{array} $

    And this cycle repeats. So we know the least residues of $\displaystyle 3^n$ (mod 13) for all $\displaystyle n$ is 3, 9, or 1.

    Since $\displaystyle a \equiv m \ (\text{mod } m)$ implies $\displaystyle m \mid a$, we have to find $\displaystyle n$ such that $\displaystyle 2^n \equiv 10, 4, \ \text{or } 12 \ (\text{mod }13)$.

    As it turns out, 2 is a primitive root modulo 13. So see for which $\displaystyle n \in [1, 12]$ such that $\displaystyle 2^n \equiv 10, 4, \ \text{or } 12 \ (\text{mod }13)$.

    See if you can figure it out from here.
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  3. #3
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    great that worked! I just determined the reside system for $\displaystyle 2^{n}$ and then checked for what values of n, their sum added appropriately!

    Merci
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