1
$\begingroup$

I have this problem from an old exam that I can't solve.

Let $\{a_i\}_{i \geq 0}$ be the series define by recursion as:

$a_0 = 2$

$a_1 = 3$

$a_{n+1} = \frac{a_n^2 +5}{a_{n-1}}$ , $\forall n \in \Bbb N$

Prove that there are integers $S$ and $T$ such that $a_{n+1} = S a_n + T a_{n-1}$

I've been trying with induction but I guess that there must be a direct, and more algebraic, way of proving it.

Any thoughts?? Thanks!

$\endgroup$
  • 1
    $\begingroup$ First things first: have you run several terms of the sequence to see what the constants $S$ and $T$ must be? $\endgroup$ – Steven Stadnicki Dec 5 '16 at 20:29
  • $\begingroup$ @Steven I've found that the constants are $S=5$ and $T=(-1)$ for $n \geq 2$. $\endgroup$ – jrs Dec 5 '16 at 20:38
  • 2
    $\begingroup$ Once you have the correct constants there's a nice characterization in terms of a matrix; note that the recursion can also be written as $a_{n+1}a_{n-1}=a_n^2+5$ or as $a_{n+1}a_{n-1}-a_na_n=5$; that form should look a lot like the determinant of a certain matrix, and you should be able to find a linear transformation that takes one matrix of that form to the same matrix with all indices shifted by one, using the recurrence relation. $\endgroup$ – Steven Stadnicki Dec 5 '16 at 20:46
  • 1
    $\begingroup$ Should S be 3 instead of 5? $\endgroup$ – displayname Dec 5 '16 at 20:58
  • 1
    $\begingroup$ Gio, it certainly does not: The next few values are $a_2=\frac{3^2+5}{2}=7$, $a_3=\frac{7^2+5}{3}=18$, and $a_4=\frac{18^2+5}{7}=47$. It's not the case, for instance, that $a_4=5a_3-a_2$. $\endgroup$ – Steven Stadnicki Dec 5 '16 at 22:14
0
$\begingroup$

Define a series $a_{n+1} = \frac{a_n^2 +k}{a_{n-1}}$ with some $a_0, a_1$ for $n\ge2$. Assume there are $S$ and $T$ such that $a_{n+1} = S a_n + T a_{n-1}$.
$$a_{n+1} = \frac{a_n^2+k}{a_{n-1}} \\ S a_n + T a_{n-1} = \frac{a_n^2+k}{a_{n-1}} \\ S a_n a_{n-1} + T a_{n-1}^2 = a_n^2+k \tag{1} $$ now $$a_{n+1} = \frac{a_n^2+k}{a_{n-1}} \\ a_{n+1}a_{n-1} = a_n^2+k \\ a_{n+1}a_{n-1} - k = a_n^2 \tag{2} $$ combine the two: $$ S a_n a_{n-1} + T(a_n a_{n-2} - k) = a_n^2+k \\ a_n(Sa_{n-1} + Ta_{n-2}) - Tk = a_n^2+k \\ a_n^2 - Tk = a_n^2+k \\ T = -1 $$ now find $S$: $$ a_{2} = \frac{a_1^2+k}{a_{0}} \\ S a_1 - a_{0} = \frac{a_1^2+k}{a_{0}} \\ S = \frac{a_0^2+a_1^2+k}{a_{0}a_{1}} \\ $$ and in this case: $$S = \frac{2^2+3^2+5}{2\times3} = \frac{18}{6} = 3$$ So $S$ is an integer if it exists. Prove by induction that $$S a_n - a_{n-1} = \frac{a_n^2+k}{a_{n-1}}$$ (note that the base case $n=2$ is covered by how we set $S$). Assume the equality is correct for all $n+1\ge k\ge3$.
$$ a_{n+2} = \frac{a_{n+1}^2+k}{a_{n}} \\ a_{n+2} = \frac{(S a_n-a_{n-1})^2+k}{a_{n}} = \frac{S^2 a_n^2-2S a_n a_{n-1}+a_{n-1}^2+k}{a_{n}} = S^2 a_n-2Sa_{n-1}+\frac{a_{n-1}^2+k}{a_{n}} \\ a_{n+2} = S(S a_n-a_{n-1})-Sa_{n-1}+\frac{a_{n-2}}{a_{n}}\frac{a_{n-1}^2+k}{a_{n-2}} = S a_{n+1} -Sa_n + \frac{a_{n-2}}{a_{n}} a_n \\ a_{n+2} = S a_{n+1} -Sa_n + a_{n-2} = S a_{n+1} -Sa_n + (S a_{n-1} - a_n) = S a_{n+1} - a_n $$ As required.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.