Demonstrate that these two Pell's equations have no integer solutions

Question

I would like to demonstrate that the following four Pell's equations have no integer solutions:

$x^2-82y^2=\pm2$

$x^2-82y^2=\pm3$

I do realise that such problems are often solved by algebraic manipulations, reducing modulo prime numbers, and arriving at some contradiction. After some blind trial and error fumbling in the dark with the above mentioned method, I have decided to consult the community.

All help or input would, as always, be highly appreciated.

Answer 1

The second one is easy $$x^2 - 82y^2 = \pm3 \Rightarrow x^2 \equiv \pm3 \pmod{41}$$ but according to Euler's criterion $$\left(\frac{\pm3}{41}\right) \equiv \left(\pm3\right)^{\frac{41-1}{2}} \pmod{41}$$ and $$\left(\pm3\right)^{20} \equiv -1 \pmod{41}$$ as a result, there is no such $x$.

Answer 2

Recommend learning the following method for the continued fraction of $\sqrt n.$ All small numbers $|k| < \sqrt n$ that have a primitive representation as $x^2 - n y^2$ show up as $p^2 - n q^2,$ where $\frac{p}{q}$ is a convergent for $\sqrt n.$ As you can see, the only small represented numbers are $\pm 1.$ See Theorem 5.1 in KCONRAD

Method described by Prof. Lubin at Continued fraction of $\sqrt{67} - 4$

$$ \sqrt { 82} = 9 + \frac{ \sqrt {82} - 9 }{ 1 } $$ $$ \frac{ 1 }{ \sqrt {82} - 9 } = \frac{ \sqrt {82} + 9 }{1 } = 18 + \frac{ \sqrt {82} - 9 }{1 } $$

Simple continued fraction tableau:
$$ \begin{array}{cccccccc} & & 9 & & 18 & & 18 & \\ \\ \frac{ 0 }{ 1 } & \frac{ 1 }{ 0 } & & \frac{ 9 }{ 1 } & & \frac{ 163 }{ 18 } \\ \\ & 1 & & -1 & & 1 \end{array} $$

$$ \begin{array}{cccc} \frac{ 1 }{ 0 } & 1^2 - 82 \cdot 0^2 = 1 & \mbox{digit} & 9 \\ \frac{ 9 }{ 1 } & 9^2 - 82 \cdot 1^2 = -1 & \mbox{digit} & 18 \\ \frac{ 163 }{ 18 } & 163^2 - 82 \cdot 18^2 = 1 & \mbox{digit} & 18 \\ \end{array} $$

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a different example:

$$ \sqrt { 229} = 15 + \frac{ \sqrt {229} - 15 }{ 1 } $$ $$ \frac{ 1 }{ \sqrt {229} - 15 } = \frac{ \sqrt {229} + 15 }{4 } = 7 + \frac{ \sqrt {229} - 13 }{4 } $$ $$ \frac{ 4 }{ \sqrt {229} - 13 } = \frac{ \sqrt {229} + 13 }{15 } = 1 + \frac{ \sqrt {229} - 2 }{15 } $$ $$ \frac{ 15 }{ \sqrt {229} - 2 } = \frac{ \sqrt {229} + 2 }{15 } = 1 + \frac{ \sqrt {229} - 13 }{15 } $$ $$ \frac{ 15 }{ \sqrt {229} - 13 } = \frac{ \sqrt {229} + 13 }{4 } = 7 + \frac{ \sqrt {229} - 15 }{4 } $$ $$ \frac{ 4 }{ \sqrt {229} - 15 } = \frac{ \sqrt {229} + 15 }{1 } = 30 + \frac{ \sqrt {229} - 15 }{1 } $$

Simple continued fraction tableau:
$$ \begin{array}{cccccccccccccccccccccccc} & & 15 & & 7 & & 1 & & 1 & & 7 & & 30 & & 7 & & 1 & & 1 & & 7 & & 30 & \\ \\ \frac{ 0 }{ 1 } & \frac{ 1 }{ 0 } & & \frac{ 15 }{ 1 } & & \frac{ 106 }{ 7 } & & \frac{ 121 }{ 8 } & & \frac{ 227 }{ 15 } & & \frac{ 1710 }{ 113 } & & \frac{ 51527 }{ 3405 } & & \frac{ 362399 }{ 23948 } & & \frac{ 413926 }{ 27353 } & & \frac{ 776325 }{ 51301 } & & \frac{ 5848201 }{ 386460 } \\ \\ & 1 & & -4 & & 15 & & -15 & & 4 & & -1 & & 4 & & -15 & & 15 & & -4 & & 1 \end{array} $$

$$ \begin{array}{cccc} \frac{ 1 }{ 0 } & 1^2 - 229 \cdot 0^2 = 1 & \mbox{digit} & 15 \\ \frac{ 15 }{ 1 } & 15^2 - 229 \cdot 1^2 = -4 & \mbox{digit} & 7 \\ \frac{ 106 }{ 7 } & 106^2 - 229 \cdot 7^2 = 15 & \mbox{digit} & 1 \\ \frac{ 121 }{ 8 } & 121^2 - 229 \cdot 8^2 = -15 & \mbox{digit} & 1 \\ \frac{ 227 }{ 15 } & 227^2 - 229 \cdot 15^2 = 4 & \mbox{digit} & 7 \\ \frac{ 1710 }{ 113 } & 1710^2 - 229 \cdot 113^2 = -1 & \mbox{digit} & 30 \\ \frac{ 51527 }{ 3405 } & 51527^2 - 229 \cdot 3405^2 = 4 & \mbox{digit} & 7 \\ \frac{ 362399 }{ 23948 } & 362399^2 - 229 \cdot 23948^2 = -15 & \mbox{digit} & 1 \\ \frac{ 413926 }{ 27353 } & 413926^2 - 229 \cdot 27353^2 = 15 & \mbox{digit} & 1 \\ \frac{ 776325 }{ 51301 } & 776325^2 - 229 \cdot 51301^2 = -4 & \mbox{digit} & 7 \\ \frac{ 5848201 }{ 386460 } & 5848201^2 - 229 \cdot 386460^2 = 1 & \mbox{digit} & 30 \\ \end{array} $$

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$$ \sqrt { 106} = 10 + \frac{ \sqrt {106} - 10 }{ 1 } $$ $$ \frac{ 1 }{ \sqrt {106} - 10 } = \frac{ \sqrt {106} + 10 }{6 } = 3 + \frac{ \sqrt {106} - 8 }{6 } $$ $$ \frac{ 6 }{ \sqrt {106} - 8 } = \frac{ \sqrt {106} + 8 }{7 } = 2 + \frac{ \sqrt {106} - 6 }{7 } $$ $$ \frac{ 7 }{ \sqrt {106} - 6 } = \frac{ \sqrt {106} + 6 }{10 } = 1 + \frac{ \sqrt {106} - 4 }{10 } $$ $$ \frac{ 10 }{ \sqrt {106} - 4 } = \frac{ \sqrt {106} + 4 }{9 } = 1 + \frac{ \sqrt {106} - 5 }{9 } $$ $$ \frac{ 9 }{ \sqrt {106} - 5 } = \frac{ \sqrt {106} + 5 }{9 } = 1 + \frac{ \sqrt {106} - 4 }{9 } $$ $$ \frac{ 9 }{ \sqrt {106} - 4 } = \frac{ \sqrt {106} + 4 }{10 } = 1 + \frac{ \sqrt {106} - 6 }{10 } $$ $$ \frac{ 10 }{ \sqrt {106} - 6 } = \frac{ \sqrt {106} + 6 }{7 } = 2 + \frac{ \sqrt {106} - 8 }{7 } $$ $$ \frac{ 7 }{ \sqrt {106} - 8 } = \frac{ \sqrt {106} + 8 }{6 } = 3 + \frac{ \sqrt {106} - 10 }{6 } $$ $$ \frac{ 6 }{ \sqrt {106} - 10 } = \frac{ \sqrt {106} + 10 }{1 } = 20 + \frac{ \sqrt {106} - 10 }{1 } $$

Simple continued fraction tableau:
$$ \begin{array}{cccccccccccccccccccccccccccccccccccccccc} & & 10 & & 3 & & 2 & & 1 & & 1 & & 1 & & 1 & & 2 & & 3 & & 20 & & 3 & & 2 & & 1 & & 1 & & 1 & & 1 & & 2 & & 3 & & 20 & \\ \\ \frac{ 0 }{ 1 } & \frac{ 1 }{ 0 } & & \frac{ 10 }{ 1 } & & \frac{ 31 }{ 3 } & & \frac{ 72 }{ 7 } & & \frac{ 103 }{ 10 } & & \frac{ 175 }{ 17 } & & \frac{ 278 }{ 27 } & & \frac{ 453 }{ 44 } & & \frac{ 1184 }{ 115 } & & \frac{ 4005 }{ 389 } & & \frac{ 81284 }{ 7895 } & & \frac{ 247857 }{ 24074 } & & \frac{ 576998 }{ 56043 } & & \frac{ 824855 }{ 80117 } & & \frac{ 1401853 }{ 136160 } & & \frac{ 2226708 }{ 216277 } & & \frac{ 3628561 }{ 352437 } & & \frac{ 9483830 }{ 921151 } & & \frac{ 32080051 }{ 3115890 } \\ \\ & 1 & & -6 & & 7 & & -10 & & 9 & & -9 & & 10 & & -7 & & 6 & & -1 & & 6 & & -7 & & 10 & & -9 & & 9 & & -10 & & 7 & & -6 & & 1 \end{array} $$

$$ \begin{array}{cccc} \frac{ 1 }{ 0 } & 1^2 - 106 \cdot 0^2 = 1 & \mbox{digit} & 10 \\ \frac{ 10 }{ 1 } & 10^2 - 106 \cdot 1^2 = -6 & \mbox{digit} & 3 \\ \frac{ 31 }{ 3 } & 31^2 - 106 \cdot 3^2 = 7 & \mbox{digit} & 2 \\ \frac{ 72 }{ 7 } & 72^2 - 106 \cdot 7^2 = -10 & \mbox{digit} & 1 \\ \frac{ 103 }{ 10 } & 103^2 - 106 \cdot 10^2 = 9 & \mbox{digit} & 1 \\ \frac{ 175 }{ 17 } & 175^2 - 106 \cdot 17^2 = -9 & \mbox{digit} & 1 \\ \frac{ 278 }{ 27 } & 278^2 - 106 \cdot 27^2 = 10 & \mbox{digit} & 1 \\ \frac{ 453 }{ 44 } & 453^2 - 106 \cdot 44^2 = -7 & \mbox{digit} & 2 \\ \frac{ 1184 }{ 115 } & 1184^2 - 106 \cdot 115^2 = 6 & \mbox{digit} & 3 \\ \frac{ 4005 }{ 389 } & 4005^2 - 106 \cdot 389^2 = -1 & \mbox{digit} & 20 \\ \frac{ 81284 }{ 7895 } & 81284^2 - 106 \cdot 7895^2 = 6 & \mbox{digit} & 3 \\ \frac{ 247857 }{ 24074 } & 247857^2 - 106 \cdot 24074^2 = -7 & \mbox{digit} & 2 \\ \frac{ 576998 }{ 56043 } & 576998^2 - 106 \cdot 56043^2 = 10 & \mbox{digit} & 1 \\ \frac{ 824855 }{ 80117 } & 824855^2 - 106 \cdot 80117^2 = -9 & \mbox{digit} & 1 \\ \frac{ 1401853 }{ 136160 } & 1401853^2 - 106 \cdot 136160^2 = 9 & \mbox{digit} & 1 \\ \frac{ 2226708 }{ 216277 } & 2226708^2 - 106 \cdot 216277^2 = -10 & \mbox{digit} & 1 \\ \frac{ 3628561 }{ 352437 } & 3628561^2 - 106 \cdot 352437^2 = 7 & \mbox{digit} & 2 \\ \frac{ 9483830 }{ 921151 } & 9483830^2 - 106 \cdot 921151^2 = -6 & \mbox{digit} & 3 \\ \frac{ 32080051 }{ 3115890 } & 32080051^2 - 106 \cdot 3115890^2 = 1 & \mbox{digit} & 20 \\ \end{array} $$

Answer 3

This is an example of a rather more general phenomenon. The continued fraction expansion of $\sqrt{t^2+1}$ is just $[t,\overline{2t}]$ and hence every convergent $p_i/q_i$ to $\sqrt{t^2+1}$ has the property that $$ p_i^2- (t^2+1) q_i^2 = \pm 1. $$ If we have $x^2-(t^2+1)y^2=k$ for a given integer $k$ and some integer $x$ with $y \neq 0$, then $$ \left| \sqrt{t^2+1} - \frac{x}{y} \right| = \frac{|k|}{y^2 \left|\sqrt{t^2+1} + \frac{x}{y} \right|} $$ and so $x/y$ is a convergent to $\sqrt{t^2+1}$ provided, roughly, $|k| \leq t$. It follows that the form $x^2-(t^2+1)y^2$ does not represent any non-square integers $k$ with $1< |k| < t$.