# Solutions of a diophantine equation

1. Mar 25, 2010

### zetafunction

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''

2. Mar 27, 2010

### Petek

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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