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Math Help - Primitive root of 2^n + 1

  1. #1
    Super Member Deadstar's Avatar
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    Primitive root of 2^n + 1

    Show that, for n>1, 3 is a primitive root of any prime of the form 2^n + 1.

    Thoughts,

    Need conditions such that if 3 is NOT a primitive root then such and such. But I don't know what the such and such should be.
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Deadstar View Post
    Show that, for n>1, 3 is a primitive root of any prime of the form 2^n + 1.

    Thoughts,

    Need conditions such that if 3 is NOT a primitive root then such and such. But I don't know what the such and such should be.
    In this case,  \phi(2^n+1) = 2^n , so  \text{ord}(3) = 2^k for some  0<k\leq n .

    Let  p=2^n+1 and look at  3^{(p-1)/2} :

     3^{(p-1)/2}\equiv \left(\frac3p\right) = \left(\frac p3\right) = \left(\frac{(-1)^n+1}{3}\right) = \left(\frac23\right) = -1\bmod{p}.
    Note that I'm saying  (-1)^n+1\equiv 2\bmod{3} since if  (-1)^n+1\equiv 0\bmod{3} then  p would not be prime.

    So in summary we have  3^{2^{n-1}}\equiv -1\bmod{p} and I claim this forces  3^{2^k}\not\equiv1\bmod{p} \;\; \forall \; k<n . Can you see why?

    Thus  \text{ord}(3) = \phi(2^n+1) which makes  3 a primitive root.
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  3. #3
    Super Member Deadstar's Avatar
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    Quote Originally Posted by chiph588@ View Post
    So in summary we have  3^{2^{n-1}}\equiv -1\bmod{p} and I claim this forces  3^{2^k}\not\equiv1\bmod{p} \;\; \forall \; k<n . Can you see why?
    Honestly, no.
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  4. #4
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by chiph588@ View Post
    So in summary we have  3^{2^{n-1}}\equiv -1\bmod{p} and I claim this forces  3^{2^k}\not\equiv1\bmod{p} \;\; \forall \; k<n . Can you see why?
    I'm saying if  3^{2^{n-1}}\equiv -1\bmod{p} , then  3^{2^k}\not\equiv1\bmod{p} \;\; \forall \; k<n .

    Let's prove the contrapositive:

    If  3^{2^k}\equiv1\bmod{p} for some  k<n , then  3^{2^{n-1}}\not\equiv -1\bmod{p} .

    Well  3^{2^{n-1}} = \left(3^{2^k}\right)^{2^{n-1-k}}\equiv 1^{2^{n-1-k}}\equiv 1\not\equiv-1\bmod{p} . Thus our original claim is true.
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  5. #5
    Super Member Deadstar's Avatar
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    Quote Originally Posted by chiph588@ View Post
    I'm saying if  3^{2^{n-1}}\equiv -1\bmod{p} , then  3^{2^k}\not\equiv1\bmod{p} \;\; \forall \; k<n .

    Let's prove the contrapositive:

    If  3^{2^k}\equiv1\bmod{p} for some  k<n , then  3^{2^{n-1}}\not\equiv -1\bmod{p} .

    Well  3^{2^{n-1}} = \left(3^{2^k}\right)^{2^{n-1-k}}\equiv 1^{2^{n-1-k}}\equiv 1\not\equiv-1\bmod{p} . Thus our original claim is true.
    Oh ok I see! I was wrongly thinking that it was either going to be \equiv to -1 or 1 for some reason. So hence we could have said that if...

    3^{\tfrac{2^{n-1}}{2}}  \equiv -1 mod p

    Then we can square both sides and get a contradiction. But of course 3^{2^k} can be congruent to more than 1 or -1...

    Cheers!
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