Pell's Equation/Examples/29

Theorem

Pell's Equation:

$x^2 - 29 y^2 = 1$

has the smallest positive integral solution:

$x = 9801$

$y = 1820$

Proof

From Continued Fraction Expansion of $\sqrt {29}$:

$\sqrt {29} = \sqbrk {5, \sequence {2, 1, 1, 2, 10} }$

The cycle is of length is $5$.

By the solution of Pell's Equation, the only solutions of $x^2 - 29 y^2 = 1$ are:

${p_{5 r} }^2 - 29 {q_{5 r} }^2 = \paren {-1}^{5 r}$

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

When $r = 1$ this gives:

${p_5}^2 - 29 {q_5}^2 = -1$

which is not the solution required.

When $r = 2$ this gives:

${p_{10} }^2 - 29 {q_{10} }^2 = 1$

From Convergents of Continued Fraction Expansion of $\sqrt {29}$:

$p_{10} = 9801$

$q_{10} = 1820$

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

$\blacksquare$

Sources

• 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $13$