# 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.

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@Brain M. Scott! thank you so much for editing – gandhi Apr 7 '12 at 10:28
Shouldn't it be $(u^2-dv^2)\color{red}{(r^2-cs^2)}=c$? – draks ... Apr 7 '12 at 11:30
Where is the difficulty? You seem to have done most of the work. You know there are infinitely many solutions to Pell's equation right? There's infinitely many $r,s$ you can use, so infinitely many pairs $(ur + dvs, us + vr)$ that solve $x^2-dy^2=c$. – Ragib Zaman Apr 7 '12 at 12:27
@gandhi : I agree with the above comment, it seems you have everything done, but before that please read FAQ in asking questions and TeX your questions properly. Some papers that can add something to it are here and here . Thank you. Try googling your answers before putting them here. It will help you and others in case of saving time. – Iyengar Apr 7 '12 at 16:35
@Ragib Zaman! I got it sir, where I was stacked. Thank you. – gandhi Apr 8 '12 at 4:44

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.

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