Pell's Equation/Examples/8
< Pell's Equation | Examples
Jump to navigation
Jump to search
Theorem
- $x^2 - 8 y^2 = 1$
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 $\sqrt 8$:
- $\sqrt 8 = \sqbrk {2, \sequence {1, 4} }$
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 $q_9$.
$\blacksquare$