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Math Help - ord p

  1. #1
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    ord p

    Let a,b be positive integers with gcd(a,b)=1.
    Suppose x^a = y^b where x,y are integers.

    Show that x = n^b and y = n^a for some integer n, using the properties of ord p .
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    What do you have so far?
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  3. #3
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    Ok so far I have no idea how to use ord p but I have made a proof which could be along the right lines....though its long here i go....

    By FTA, we can express x,y as a product of primes uniquely:

    x = (p1)^s1 x (p2)^s2...(pn)^sn
    y = (p1)^t1 x (p2)^t2...(pn)^tn

    Then x^a = (p1)^as1 x (p2)^as2...(pn)^asn
    and y^b = (p1)^bt1 x (p2)^bt2...(pn)^btn

    Since x^a = y^b,
    (p1)^as1 x (p2)^as2...(pn)^asn = (p1)^bt1 x (p2)^bt2...(pn)^btn

    => asi = bti for all 1<= i <= n

    so a divides bti and b divides asi.

    Since gcd(a,b)=1, by Euler's lemma, a divides ti and b divides si.
    => ti = azi and si = byi fo some integer zi, yi

    => x^a = (p1)^aby1 x (p2)^aby2...(pn)^abyn
    y^b = (p1)^abz1 x (p2)^abz2...(pn)^abzn

    Again since x^a = y^b,
    (p1)^aby1 x (p2)^aby2...(pn)^abyn =(p1)^abz1 x (p2)^abz2..(pn)^abzn

    => yi = zi for all 1<= i<= n

    So x^a = (p1)^abz1 x (p2)^abz2...(pn)^abzn = [(p1)^z1 x (p2)^z2...(pn)^zn]^ab

    => x = [(p1)^z1 x (p2)^z2...(pn)^zn]^b

    Similarly y^b = [(p1)^z1 x (p2)^z2...(pn)^zn]^ab
    => y = [(p1)^z1 x (p2)^z2...(pn)^zn]^a

    Let n be some integer s.t. n = [(p1)^z1 x (p2)^z2...(pn)^zn]
    Then we get x = n^b and y = n^a.


    But do I have to worry about signs?? What about ord p? Is it possible you can give an argument as to how you would go about it??
    Thanks!
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  4. #4
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    Sorry to pester you, but any comments on what I have done? Thanks
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  5. #5
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by vinnie100 View Post
    Sorry to pester you, but any comments on what I have done? Thanks
    This arguement looks fine to me. I don't know how you would let order modulo p come into play...

    For the sign, I would say
    x = (-1)^c x (p1)^s1 x (p2)^s2...(pn)^sn
    y = (-1)^d x (p1)^t1 x (p2)^t2...(pn)^tn
    and go about your arguement as before.
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  6. #6
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    Thanks very much for confirming this. Ok I will just ignore the ord p
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