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Solutions of a diophantine equation

  1. Mar 25, 2010 #1
    given the diophantine polynomial equation

    [tex] f(x)=0mod(p) [/tex]

    then is the number of solution approximately less than a given N approximately

    [tex] \sum_{i\le N}e^{2i p\pi f(j)} [/tex]

    the idea is that the sum takes its maximum value every time p divides f(j) for some integer 'j''
     
  2. jcsd
  3. Mar 27, 2010 #2
    I replied yesterday (March 26), but the reply was lost with the server problems. I'll try again:

    Your summation expression appears to have a typo. If n is any integer, then

    [tex]e^{2i p\pi n} = 1[/tex]

    so the expression always sums to N. Perhaps you meant to divide by p in the exponent instead of multiplying by p. Questions of this sort are discussed in the first few sections of Number Theory by Borevich and Shafarevich.

    Additional comments:

    1. You might want to test your formula with [itex]f(x) = x^p - x[/itex], since all natural numbers are solutions.

    2. In general, your sum will be a complex number. In what sense do you want to consider a complex number to approximate the number of solutions?

    Petek
     
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