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.
 A: 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 $n \in \mathbb{Z}$.
A: For proof, you can use the solutions of the equation Pell:  $x^2-dy^2=1$
If the ratio is not a square, then the answer is always there. $(x_0;y_0)$
If we have any solution of the equation Pell: $x^2-dy^2=c$ 
Having a form: $(x_1;y_1)$
The following solution we can always be obtained by the formula:
$x_2=x_0x_1+dy_0y_1$
$y_2=y_0x_1+x_0y_1$
Obtained these values should be substituted back into the formula. And so this process can continue indefinitely. And we will get an infinite number of solutions.
A: Thought it possible to simplify in order to be able to write the solutions of the equation. For this we use the decomposition of the number $c$ on the multipliers. 
$$Z^2-dR^2=c=ab$$
To record decisions have to know first the solution of the Pell equation $(Z_1;R_1)$.
And solving the following equation Pell $(k_0;n_0)$.
$$k^2-dn^2=1$$
Then the formula is as follows.
$$Z_2=k_0Z_1+dn_0R_1$$
$$R_2=n_0Z_1+k_0R_1$$
The problem in finding the first solution for General Pell equation $(Z_1;R_1)$.
The meaning of the solution is that to factor the number. $c=ab$
Then degradable factoring the difference.  $xy=a-b$
If the following expression may be a square.  
$$s^2=\frac{1}{d}((\frac{y+x}{2})^2-a)$$
Then the first solution is written simply.
$$Z_1=ds^2+\frac{y^2-x^2}{4}$$
$$R_1=ys$$
Such record these formulas will greatly simplify the calculations. Always better to have a formula.
