Question

Let $\alpha and\beta$ be the roots of the quadratic equation $x^2+mx-1=0$, where m is an odd integer. Let ${\lambda }_n={\alpha }^n+{\beta }^n$, for $n\ge 0$. Prove that for $n\ge 0$.
(a) ${\lambda }_n$ is an integer; and
(b) $GCD({\lambda }_n,{\lambda }_{n+1})=1$

Answer 1

(1) Using mathematical induction
${\lambda }_0={\propto }^0+{\beta }^0=2\to integer.$
${\lambda }_1=\propto +\beta =-m=integer.$
${\propto }_2-{\propto }^2+{\beta }^2=({\propto }^2+{\beta }^2)-2\propto \beta =$ integer since $\propto +\beta \&\propto \beta$ are integers.
Assume ${\lambda }_n$is an integer
${\lambda }_{n+1}={\lambda }^{n+1}+{\beta }^{n+1}=({\propto }^n+{\beta }^n)(\lambda +\beta )-\lambda \beta ({\lambda }^{n-1}+{\beta }^{n-1})$
$=-m\left({\lambda }_n\right)+{\lambda }_{n+1}$
& proceeding to ${\lambda }_2,\lambda ,{\lambda }_0\to$ R.H.S. is an integer.
${\lambda }_{n+1}$ is an integer.
$\Rightarrow {\lambda }_n$ is an integer.
(2) ${\lambda }_{n+1}.{\lambda }_n(\propto +\beta )-\propto \beta {\lambda }_{n-1}$ - (1)
Now we know ${\lambda }_0\&{\lambda }_1$ are co - primes.
Similarly ${\lambda }_1\&{\lambda }_2$ are also co - prime
Assume, ${\lambda }_n\&{\lambda }_{n-1}$ are relatively prime.
Taking (1), if d divides ${\lambda }_{n+1}\&{\lambda }_n$, then d will divide $\lambda \beta {\propto }_{n-1}.But\propto \beta =-1\therefore ddivides{\lambda }_{n-1}Butwe{\lambda }_n\&{\lambda }_{n-1}$ are relatively prim. Hence d cannot divide ${\lambda }_{n-1}$. This is contradicting induction assumption.
$\therefore GCD({\lambda }_n,{\lambda }_{n+1})=1$