Pell's Equation/Examples/8

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Theorem

Pell's Equation:

$x^2 - 8 y^2 = 1$

has the positive integral solutions:

\(\ds \tuple {x, y}\) \(=\) \(\ds \tuple {3, 1}\)
\(\ds \tuple {x, y}\) \(=\) \(\ds \tuple {17, 6}\)
\(\ds \tuple {x, y}\) \(=\) \(\ds \tuple {99, 35}\)
\(\ds \tuple {x, y}\) \(=\) \(\ds \tuple {577, 204}\)
\(\ds \tuple {x, y}\) \(=\) \(\ds \tuple {3363, 1189}\)

and so on.


Proof

From Continued Fraction Expansion of $\sqrt 8$:

$\sqrt 8 = \sqbrk {2, \sequence {1, 4} }$

The cycle is of length $2$.

By the solution of Pell's Equation, the only solutions of $x^2 - 8 y^2 = 1$ are:

${p_{2 r} }^2 - 8 {q_{2 r} }^2 = \paren {-1}^{2 r}$

for $r = 1, 2, 3, \ldots$

When $r = 1$ this gives:

${p_2}^2 - 8 {q_2}^2 = 1$

which is the solution required.

From Convergents of Continued Fraction Expansion of $\sqrt 8$:

$p_2 = 3$
$q_2 = 1$

although on that page the numbering goes from $p_0$ to $p_9$, and $q_0$ to $q_9$.

$\blacksquare$