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Find Solutions- Number Theory

  1. Apr 26, 2007 #1

    Gib Z

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    This isn't homework, but an interest question I found on the web. I can't solve it though...

    Let [itex]p[/itex] be an odd prime. Determine all pairs [itex](m,n)[/itex] where m and n are positive integers and satisfy the below:

    [tex](p-1)(p^n+1)=4m(m+1)[/tex].

    I have done by some simple inspection that m must be odd, but I'm not sure that helps. Does anyone know any theorems that may help?
     
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  3. Apr 26, 2007 #2

    Gib Z

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    I have made a tiny bit of progress I think.

    I Make it a quadratic and solved:
    [tex]m=\frac{-1\pm \sqrt{1-(p-1)(p^n+1)}}{2}[/tex].

    Since m must be odd, 2m must be even. So the numerator must be even. So the square root must be odd. But for m to be an integer, the square root must also be an integer. So I get [itex]1-(p-1)(p^n+1)[/itex] must be a perfect square. How Do i continue?


    EDIT: Looking again...that can't be right because (p-1)(p^n+1) must be less than one, or exactly 1. It can't be less than one for obvious reasons, and it can't be exactly 1 either.

    EDIT2 : Solved quadratic wrong..Ill try again
     
    Last edited: Apr 26, 2007
  4. Apr 26, 2007 #3

    Gib Z

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    Ok so the actual solution to the quadratic is
    [tex]m=\frac{-1\pm \sqrt{1+(p-1)(p^n+1)}}{2}[/tex], sign error...

    Well the numerator must be even, so the square root must be odd. So to be an integer, 1+(p-1)(p^n+1) must be the square of an odd number, so of the form (2k+1)^2 where k is some integer. So (p-1)(p^n+1)=4k^2+4k, so (p-1)(p^n+1) must be divisible by 4.

    Can anyone help me from here?
     
  5. Apr 26, 2007 #4

    Gib Z

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    I Just realised now that I did Nothing. Nothing at all. The result I achieved can be done by dividing the original questions equation by 4...
     
  6. Apr 26, 2007 #5

    Office_Shredder

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    Alternatively, p-1 is even, p^n + 1 is even, so it must be divisible by four. So the equation doesn't even force the issue, just the way that (p-1)(p^n + 1) works
     
  7. Apr 26, 2007 #6
    where did u find it?
     
  8. Apr 26, 2007 #7

    Gib Z

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    It was on some other forums but no solution was found there either, and the thread was closed for some reason..
     
  9. Apr 26, 2007 #8

    Office_Shredder

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    Well, solutions do exist (if p=5, m=2 n=1). Interestingly, this disproves your m is odd claim (which seemed, well... odd to me, considering m+1 would always be even then).
     
  10. Apr 26, 2007 #9
    how is m always odd?

    [tex](p-1)(p^n+1)=4m(m+1)[/tex]

    if p is odd, then p - 1 is even. since p^n is always odd (i checked it using a C++ program :) ), then the expression p^n - 1 is even. even times even is even. so, in order for the whole thing to be equal, the right side must be even. if you put m as even, then you get (even)*(odd) which equals even.If you put m as odd, then you still get an even right side.
     
  11. Apr 27, 2007 #10
    I sometimes wonder whether the poster is playing games with his posts. Anyway, a solution for every odd prime is n = 1 and m = (p-1)/2 so m can be both odd or even. In fact this solution works for every odd composite p as well, but that is precluded in the statement of the problem.
     
  12. May 4, 2007 #11

    Gib Z

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    Jesus Christ..sorry I haven't replied guys But I worked out the full solution in case anyones interested :)

    [tex](p-1)(p^n+1)=4m(m+1)[/tex]
    [tex]p^{n+1}-p^n+p = 4m(m+1) + 1 = (2m+1)^2[/tex]

    Let k be some odd number, since of the form 2m+1

    [tex]]p^{n+1}-p^n+p=k^2[/tex]

    For n=1, solving for m gives the solution given by ramsey. For n > 1, assume it has solutions. Then P divides K. Then divide both sides by p^2. That gives p^(n-1) - p^(n-3) + 1/(p) = k^2/(p^2) = (k/p)^2

    Since it is implied P divides k, the RHS should be an integer, but the LHS is not. Done.
     
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