We would like to solve the diophantine
$$7x^2-5y^2=18 \tag{E}$$
We first solve the linear diophantine
$$7x-5y=18$$
Solutions are couples $(7k+2,5k+4)$ where $k$ is an arbitrary integer.
thus $(x,y)$ satisfies (E) iff $x^2 \equiv 2 \pmod 7$ and $y^2\equiv 4 \pmod 5$
Hence $x\equiv 3 , 4 \pmod 7$ and $y\equiv 2 ,3\pmod 5$
i wonder how to continue from here.
thanks for your time.
The trick here is to multiply your equation by 7 to obtain $$(7x)^2 - 35y^2 = 126.$$ From here you can first look at the equation $x^2 - 35y^2 = 126$, and then divide $x$ by $7$ to find the solutions of your original equation (note that if $(x, y)$ is a solution, then necessarily $x$ is divisible by $7$). This equation can be solved by the general procedure for Pell equations: you first find the smallest solution, and all other solutions are generated by solutions of $x^2 - 35y^2 = 1$.
Edit: You get the full set of solutions of $x^2 - 35y^2 = 126$ by starting with the solution $(x_0, y_0) = (21, 3)$, and then the smallest solution of $x^2 - 35y^2 = 1$ is $(x, y) = (6, 1)$, so you can build recursively solutions $(x_{n+1}, y_{n+1}) = (6x_n + 35y_n, x_n + 6y_n)$, and in order to find the solutions of your original equation you need to divide $x$ by $7$, which gives you $(x_0, y_0) = (3, 3)$ and $(x_{n+1}, y_{n+1}) = (6x_n + 5y_n, 7x_n + 6y_n)$, and the general theory of Pell equations tells you that these are the only solutions (at least with $x, y$ positive. If you change the sign of $x$ or $y$ it is obviously also a solution).