Solution

From part (b),

a_{2 m + 1} = 3

and

a_{2 m + 2} = 6

. Eliminating the intervening odd-indexed convergent from the standard recurrence shows that either sequence

x_{m} = p_{2 m}

x_{m} = q_{2 m}

obeys

x_{m + 2} = 20 x_{m + 1} - x_{m} .

The initial even convergents are

\frac{p _{0}}{q _{0}} = \frac{3}{1}, \frac{p _{2}}{q _{2}} = \frac{63}{19} .

Put

u = 10 + 311

. Since

N (u) = 100 - 99 = 1

and

u + u^{- 1} = 20

, the sequence

w_{m} = (3 + 11) u^{m}

satisfies the same recurrence. Its first two values are

3 + 11

and

63 + 1911

, so

p_{2 m} + q_{2 m} 11 = (3 + 11) (10 + 311)^{m} .

Taking norms in

Q (11)

yields

p_{2 m}^{2} - 11 q_{2 m}^{2} = (9 - 11) (100 - 99)^{m} = - 2 .

This applies in particular whenever the original index

n = 2 m \geq 2

is even.

Solved by gpt-5.6-sol high.

 Ancestors (11)