Mysterious formulae involving the numbers of points in some families of elliptic curves
The project is to find a (good) explanation of the following experimental fact. for each prime p at least five, we look at the weierstrass equations y^{2} = 4x^{3}  g_{2}x  g_{3} with coefficients g_{2} and g_{3} in the prime field F_{p} and for which the discriminant (g_{2})^{3}  27(g_{3})^{2} is equal to ONE. For each such elliptic curve, its number of F_{p} points (including the point at infinity) we write as 1 + p  A(p, g_{2}, g_{3}). Then we form the sum of A(p, g_{2}, g_{3}) over all the g_{2} and g_{3} in the prime field F_{p} with discrimint ONE, call this sum B(p). okay, it is not hard to show that for prime p which are 3 mod 4, we have B(p) = 0. And one finds experimentally that for a prime p which is 1 mod 4, if we write p as the sum of two squares a^{2} + b^{2} with a and b both positive and with a ODD, then we have the miraculous formula
B(p) = 2(a^{2}  b^{2}).
How come?
Here is a similar fact, in this case provably true, but even here I don't know a really satisfying explanation. One takes, for each prime p at least three, all the modified legendre equations y^{2} = a(a1)x(x1)(xa) over F_{p}, now with a in F_{p} but not 0 or 1. one writes the number of F_{p} points on this curve as 1 + p  A(p,a), then one forms the sum of A(p,a) over all the a not 0 or 1 in F_{p}, call this sum B(p). The notation is not as bad as it seems, because this B(p) is 0 for primes p which are 3 mod 4, and if p is 1 mod 4, if we write p as the sum of two squares a^{2} + b^{2} with a and b both positive and with a ODD, then we have the SAME formula
B(p) = 2(a^{2}  b^{2}).
How come? And why is "this" B(p) equal to the one above? In other words, can one see that the two B(p)'s are equal to each other without knowing what their common numerical value is? How can one predict such phenomena?
good luck, nick katz
