# Solutions to x^2 – dy^2 = 1 and x^2 – dy^2 = -1

Recall from the Pell’s Equation page that Pell’s equation is $x^2 – dy^2 = N$ where $d, N \in \mathbb{Z}$. We noted that Pell’s equation has at most finitely many solutions if $d or $d$ is a perfect square. We will now look at a special type of Pell’s equation and when it has solutions, namely:

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## Solutions to x^2 – dy^2 = 1

Proposition 1: If $d > 0$ is not a perfect square and $(a, b)$ is a positive solution to $x^2 – dy^2 = 1$, i.e., $a, b > 0$ then $x^2 – dy^2 = 1$ has infinitely many solutions. |

**Proof:**Note that $x^2 – dy^2 = 1$ has the solution $(x, y) = (1, 0)$. Let $(a, b)$ be any solution to $x^2 – dy^2 = 1$. Then:

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- Squaring both sides of equation above yields:

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- So $(x, y) = (a^2 + b^2d, 2ab)$ is another solution to $x^2 – dy^2 = 1$. $\blacksquare$

Proposition 2: If $d > 0$ is not a perfect square and $\frac{h_n}{k_n}$ is the $n^{\mathrm{th}}$ convergent of $\sqrt{d}$ then $h_n^2 – dk_n^2 = (-1)^{n-1} q_{n+1}$. |

*Recall that $\displaystyle{q_{n+1} = \frac{d – m_{n+1}^2}{q_n}}$ and $m_{n+1} = a_nq_n – m_n$.*

Corollary 3: Suppose that $d > 0$ and $d$ is not a perfect square. Let $r$ be the period of the continued fraction expansion of $\sqrt{d}$. Then $(\pm h_{rn – 1}, \pm k_{rn – 1})$ satisfies the equation $x^2 – dy^2 = (-1)^{rn}$. |

Corollary 4: If $d > 0$ is not perfect square then $x^2 – dy^2 = 1$ has a positive solution and hence infinitely many solutions. |

**Proof:**By corollary $3$, for every even $n$ we have that $nr$ is even, and so $(\pm h_{rn – 1}, \pm k_{rn – 1})$ is a solution to $x^2 – dy^2 = 1$ for every even $n$.

- Alternatively, we can generate a single positive solution by corollary 3 and then apply proposition 1. $\blacksquare$

Theorem 5: Let $d > 0$ and $d$ not be a perfect square. If $(x_1, y_1)$ is the first positive solution to $x^2 – dy^2 = 1$ then all positive solutions to $x^2 – dy^2 = 1$ are of the form $x_n + y_n \sqrt{d} = (x_1 + y_1 \sqrt{d})^n$. |

## Solutions to x^2 – dy^2 = -1

From corollary 3, if $r$ is even then $rn$ is always even and so $x^2 – dy^2 = -1$ cannot have solutions since $(-1)^{rn} = 1$. If $r$ is odd then $rn$ is odd infinitely often and so $x^2 – dy^2 = -1$ has infinitely many solutions by corollary 3. We summarize this below.

Corollary 6: Suppose that $d > 0$ and $d$ is not a perfect square. Let $r$ be the period of the continued fraction expansion of $\sqrt{d}$. If $r$ is odd then the equation $x^2 – dy^2 = -1$ has infinitely many solutions, namely $\left (\pm h_{rn – 1}, \pm k_{rn – 1} \right )$ are solutions. |

For example, consider the Pell’s equation:

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The continued fraction expansion of $\sqrt{d} = \sqrt{13}$ is:

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So $r = 5$, which is odd. Corollary 2 says that $x^2 – 13y^2 = -1$ has solutions and furthermore, $(\pm h_{5r – 1}, k_{5n – 1} )$ are solutions for every $n$. If we see $n = 1$ then we see that $(h_4, k_4)$ is a particular solution. Let’s verify this.

$n$ | $a_n$ | $h_n$ | $k_n$ |
---|---|---|---|

$-2$ | – | $0$ | $1$ |

$-1$ | – | $1$ | $0$ |

$0$ | $3$ | $3$ | $1$ |

$1$ | $1$ | $4$ | $1$ |

$2$ | $1$ | $7$ | $2$ |

$3$ | $1$ | $11$ | $3$ |

$4$ | $1$ | $18$ | $5$ |

So $(h_4, k_4) = (18, 5)$. Note that indeed

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