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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.

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    What if $d = 1$? Then $x^2-y^2=2n+1$ has only a finite number of solutions for any integer $n$ and it always has the solution $x = n+1$, $y = n$.2014-06-12

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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 $n \in \mathbb{Z}$.

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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.

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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.