# How many different pairs of integers $(x,y)$ modulo $p$ such that $ax^2+by^2+c \equiv 0\pmod{p}$?

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)

• This question seems to be related with this one here: math.stackexchange.com/questions/1190423/… – Dr. Mathva Nov 26 '18 at 18:38
• @Dr.Mathva Yes, thank you. However I am looking for the number of solutions, not just if a solution exists. – color Nov 26 '18 at 18:43
• Let $\chi(n) = n^{(p-1)/2}= \pm 1, \chi(0) = 0$. Then there are $\sum_{x \bmod p} \chi(b^{-1})\chi(-ax^2-c)+\chi(b^{-1})^2\chi(-ax^2-c)^2$ solutions with $y \ne 0$. Can you go further assuming $-ac^{-1} = d^2$ ? – reuns Nov 27 '18 at 0:56

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.