# Exponential sums and congruences

1. Oct 27, 2008

### mhill

let be the exponential sum

$$S= \sum_{n=1}^{N}e( \frac{f(x)}{p})$$

$$e(x)= exp( 2i \pi x)$$

my conjecture is that since the complex exponential takes its maximum value '1' when x is equal to an integer then

$$Re(S)= \Pi (f,N)$$ with $$\Pi (f,N)$$ is the number of solutions on the interval (1,N) of the congruence

$$f(x) =0 mod(p)$$ and f(x) is a Polynomial.

2. Nov 4, 2008

### yasiru89

Forgive me if this is a stupid question- but what's [itex]p[/tex]? Or did you mean [itex]n[/tex] instead or [itex]p[/tex] as the number of prime factors of [itex]n[/tex] or something?

3. Nov 5, 2008

any prime

4. Nov 5, 2008

### yasiru89

I still don't get over what exactly that summation for [itex]S[/tex] is done. A clarification please?

5. Nov 6, 2008

### mhill

o sorry.. i should have written

$$S= \sum_{n=1}^{N}e( \frac{f(n)}{p})$$

the sum is taken over 'n' but if the prime 'p' divides f(n) then the complex exponential is equal to '1'

6. Nov 6, 2008

### yasiru89

Okay then [itex]p[/tex] is a prime of one's choosing.

We have,

$$S = \sum_{n = 1}^{N} \exp{\left(\frac{2\pi i}{p}f(n)\right)}$$

Then,
$$\Re(S) = \sum_{n = 1}^{N} \cos{\left(\frac{2\pi}{p}f(n)\right)}$$

If [itex]f(n)[/tex] is a multiple of [itex]p[/tex], then the the real part of [itex]S[/tex] will 'count' each solution of that congruence, but what about certain [itex]f(n)[/tex] values that don't and hence give rise to non-zero real and imaginary components? They won't be 1 in a single go, but they can possibly accumulate to values greater than 1 I think. So some bounds for such a theorem also become necessary if I haven't missed anything.

Last edited: Nov 6, 2008