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Number of theory modulo

  1. Oct 17, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the smallest natural number n such that: a^(3n)=a (mod 85) for each integer a.
    Justify your answer.

    2. Relevant equations


    3. The attempt at a solution

    because 85=17 . 5 and gcd (5,17)=1 we have to find the n such:

    a^(3n)=a (mod 5) and a^(3n)=a (mod 17).

    From Fermat theorem we know that a^(17)=a (mod 17) and a^(5)=a (mod 5)

    so we have: a^(5)=a^(3n) (mod 5) and a^(17)=a^(3n)(mod 17).

    I don't now how to continue and find the smallest natural n.
     
  2. jcsd
  3. Oct 17, 2012 #2
    You've already observed that it suffices to find a minimal n such that [itex]a^{3n} \equiv a \pmod{5}[/itex] and [itex]a^{3n} \equiv a \pmod{17}[/itex]. Here's a hint about how to do so:

    You already know that [itex]a^{5} \equiv a \pmod{5}[/itex]. Show that [itex]a^9 \equiv a \pmod{5}, a^{13} \equiv a \pmod{5}[/itex] and so on.

    Similarly, you know that [itex]a^{17} \equiv a \pmod{17}[/itex]. Show that [itex]a^{33} \equiv a \pmod{17}[/itex], and so on.

    Use the above to find the least exponent 3n such that [itex]a^{3n} \equiv a \pmod{5}[/itex] and [itex]a^{3n} \equiv a \pmod{17}[/itex].

    Please post again if you have any questions.
     
  4. Oct 18, 2012 #3
    i just do it. i find a^(33)=a(mod5) and a^(33)=a(mod17)
    now can i say that 3n=33 and n=11. is that right ???
     
  5. Oct 18, 2012 #4
    I got the same answer.
     
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