Let $p$ be an odd prime number and $a,b,c$ be integers coprime with $p$. How many different pairs of integers $(x,y)$ modulo $p$ such that $ax^2+by^2+c \equiv 0\pmod{p}$ ?
Until now I haven't had any specific way to approach this problem. How can I find the number of solutions of the equation $ax^2+by^2+c \equiv 0\pmod{p}$ ? Or can it only be counted with some given conditions of $a,b,c$ ?
(Please let me know if I should add some details to this problem)
Suppose There is a number d which is prime to p such that we can write following relations (for a simple form let $c=-1$):
$d^{p-1}-1 ≡0\mod p$
$ax^2+by^2=d^{p-1}$
If $y=0$ then:
$$x=\frac{d^{(p-1)/2}}{a^{1/2}}$$
So
$x=<\frac{d^{(p-1)/2}}{a^{1/2}}$
That is $1,2,3, . . \frac{d^{(p-1)/2}}{a^{1/2}}$ can be the solutions for x, so number of solutions for x is $$\chi_x=\frac{d^{(p-1)/2}}{a^{1/2}}$$ . Clearly there will be corresponding value for y for each value of x. The equation is symmetric and the same argument could be applied for y, therefore:
$$y=<\frac{d^{(p-1)/2}}{b^{1/2}}$$
That is the number of solutions for y is $$\chi_y=\frac{d^{(p-1)/2}}{b^{1/2}}$$ So the number of solution must be equal to:
$$\chi=\frac{d^{(p-1)/2}}{a^{1/2}}+\frac{d^{(p-1)/2}}{b^{1/2}}$$
You can exclude 1 from solutions and find better formula.