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Topic: discriminant of quadratic field relating to a modular sum
Replies: 12   Last Post: Oct 3, 2013 5:32 AM

 Messages: [ Previous | Next ]
 Ego Zim Posts: 24 Registered: 11/10/07
discriminant of quadratic field relating to a modular sum
Posted: Sep 25, 2013 6:10 PM

Hi,

Let G(x,y) = 2 * (x^3 * y^3)

Let M_1(n) = the sum of [x from 1 to (n-1)] of the sum of [y from 1 to (n-1)] of G(x,y)

Let M_2(n) = the sum of [x from 1 to (n-1)] of the sum of [y from 1 to (n-1)] of (G(x,y) mod n)

Let D(n) = Negative of the discriminant of quadratic field Q(sqrt(-n)) http://oeis.org/A204993

I've noticed that 4*M_1(n) divides D(n-1) without remainder and also 2*M_2(n) mod n divides D(n-1) without remainder.

A gracious gratitude and eternal blessing is guaranteed to the brave soul that can explain this, and possibly prove it.

Here is a simple c# code to illustrate:

int[] QField = new int[] { 4, 8, 3, 4, 20, 24, 7, 8, 4, 40, 11, 3, 52, 56, 15, 4, 68, 8, 19, 20, 84, 88, 23, 24, 4, 104, 3, 7, 116, 120, 31, 8, 132, 136, 35, 4, 148, 152, 39, 40, 164, 168, 43, 11, 20, 184, 47, 3, 4, 8, 51, 52, 212, 24, 55, 56, 228, 232, 59, 15, 244, 248, 7, 4, 260 };

for (double N = 2; N < 65; N++)
{
double Sum = 0;
for (double y = 1; y < N; y++)
for (double x = 1; x < N; x++)
Sum += (Math.Pow(x, 3) * Math.Pow(y, 3) * 2) % N;

// always integer
double Result = 2 * Sum / QField[(int)N - 1];
}

Thanks,
Ofer

Date Subject Author
9/25/13 Ego Zim
9/25/13 Ego Zim
9/25/13 James Waldby
9/26/13 Ego Zim
9/26/13 quasi
9/26/13 Ego Zim
9/26/13 quasi
9/27/13 quasi
9/28/13 Ego Zim
9/28/13 quasi
9/28/13 quasi
9/28/13 Ego Zim
10/3/13 Leon Aigret