View Full Version : solutions of a diophantine equation
zetafunction
Mar25-10, 04:13 PM
given the diophantine polynomial equation
f(x)=0mod(p)
then is the number of solution approximately less than a given N approximately
\sum_{i\le N}e^{2i p\pi f(j)}
the idea is that the sum takes its maximum value every time p divides f(j) for some integer 'j''
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
e^{2i p\pi n} = 1
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 f(x) = x^p - x, 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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