Whenever Pell's equation proof is solvable, it has infinitely many solutions

Prove that whenever the equation $x^2 - dy^2 = c$ is solvable, then it has infinitely many solutions.

I consider that, if $u$ and $v$ satisfy $x^2 -dy^2 = c$ and then $r$ and $s$ satisfy $x^2 -cy^2 = 1$, then $$(ur \pm dvs)^2 - d(us \pm vr)^2 = (u^2 - dv^2)(r^2 - ds^2) = c\;.$$ But, still I failed to complete the proof. I am requesting members to spare some time for this. Thanks in advnace.

Pell's equation $x^2 - d y^2 = 1$ always has a fundamental solution $(x_0, y_0)$ (solution with smallest $x > 1$). All other solutions can be expressed: $$ x_n - y_n \sqrt{d} = (x_0 - y_0 \sqrt{d})^n $$ It so happens that if you define the norm in the ring $\mathbb{Z}(\sqrt{d}) = \{a + b \sqrt{d} \colon a, b \in \mathbb{Z}\}$ by: $$ N(a + b \sqrt{d}) = a^2 - b^2 d $$ then if you define the conjugate of $z = x + y \sqrt{d}$ by $\overline{z} = x - y \sqrt{d}$ you have: $$ N(z) = N(\overline{z}) = z \cdot \overline{z} $$ Also, since $\overline{u \cdot v} = \overline{u} \cdot \overline{v}$, it is also: $$ N(u \cdot v) = (u \cdot v) \cdot (\overline{u \cdot v}) = (u \cdot \overline{u}) \cdot (v \cdot \overline{v}) = N(u) \cdot N(v) $$ Your given solution is $N(r - s \sqrt{d}) = c$, and we have $N(x_0 - y_0 \sqrt{d}) = 1$: $$ N((r - s \sqrt{d}) \cdot (x_0 - y_0 \sqrt{d})^{\pm n}) = N(r - s \sqrt{d}) \cdot \left(N(x_0 - y_0 \sqrt{d})\right)^{\pm n} = c $$ I.e., $(r - s \sqrt{d}) \cdot (x_0 - y_0 \sqrt{d})^n$ defines a solution for all $n \in \mathbb{Z}$.