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Congruence identities using Fermat's Little Theorem

  1. Apr 15, 2012 #1
    1. The problem statement, all variables and given/known data

    Show the remainder when 43^43 is divided by 17.

    2. Relevant equations

    $$43 = 16 \times 2 + 11$$
    $$a^{p-1}\equiv1\ (mod\ p)$$

    3. The attempt at a solution

    I believe I can state at the outset that as:
    $$43\equiv9\ (mod\ 17)$$
    Then
    $$43^{43}\equiv9^{43}\ (mod\ 17)$$
    and that I can rewrite this as:
    $$9^{43}=\left(9^{2}\right)^{16}\times9^{11}$$
    Then applying Fermat's Little Theorem we get:
    $$\left(9^{2}\right)^{16}\equiv1\ (mod\ 17)$$
    So that
    $$43^{43}\equiv 1\times9^{11}\ (mod\ 17)$$
    But I'm unsure where to go from here.
    Any help appreciated
     
  2. jcsd
  3. Apr 15, 2012 #2
    There are numerous ways to go about solving the problem from this kind of point. You just try to simplify the numbers more and more. A nice way I found to solve this is to consider [itex]9^2 (mod 17)[/itex]. You can simplify the problem nicely in the next step after that.
     
  4. Apr 15, 2012 #3
    Thanks for taking the time out.
    I feel like I'm probably not far away!

    $$9^2\equiv13\ (mod\ 17)$$

    So I guess I can consider $$9^{11}=\left(9^2\right)^5\times9$$ and simplify to:

    $$43^{43}\equiv13^5\times 9$$

    But still unsure how to proceed. I'll have another think...
     
  5. Apr 15, 2012 #4
    Okay, good! What is another way of writing 13 mod 17?
     
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