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I'm having problems with this exercise. Let $k > 1$ be an integer. We define $(a_n)_{n \in \Bbb N_0}$ as: $$a_0 = 1$$ $$a_1 = 1$$ $$a_{n+2} = (k^2-2)a_{n+1}-a_n-2(k-2)$$ Prove that $\forall n \in \Bbb N_0, a_n$ is a perfect square.

I'm not sure how to tackle this problem. I want to prove that $a_n=(a+b)^2$ in some way. I was going to attempt with induction, $P(0)$ and $P(1)$ are true, $P(2)=(k-1)^2$... but how should I pick $P(n)$ and $P(n+1)$. Or should I try other method?

I've also tried finding the roots so I could have the closed formula for the sequence but I've ended up with a rather disappointing looking expression: $\frac{(k^2-2) \pm \sqrt{k^4-4k^2-8k+16}}{2}$

Any ideas or suggestions? Thanks!

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  • $\begingroup$ Have you tried explicitly computing $P(3)$, $P(4)$, etc. to see if you can find a pattern? $\endgroup$ – Steven Stadnicki May 2 '16 at 0:33
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    $\begingroup$ THat screams induction to me. An the set up is right there. All you have to do is hit it with a paddle. Prove if $(k^2-2)a_{n+1} - a_n - 2(k - 2)=m^2$ is a perfect square then $((k+1)^2 - 2)(k^-2)a_{n+1} - a_n - 2(k - 2) - a_{n+1} - 2(k-1)= ((k+1)^2 - 2)m^2 -j^2- 2k - 2$ is also a perfect square. $\endgroup$ – fleablood May 2 '16 at 0:49
  • $\begingroup$ @fleablood, be interesting to see a really easy treatment of this. $\endgroup$ – Will Jagy May 2 '16 at 1:53
  • $\begingroup$ Gio, where did you get this problem? $\endgroup$ – Will Jagy May 2 '16 at 1:55
  • $\begingroup$ @Will it's from an algebra course of the university of buenos aires. It's the number 24 cms.dm.uba.ar/academico/materias/1ercuat2016/algebra_I/… (it's in spanish). I will provide more context in future posts thanks for the help. $\endgroup$ – jrs May 2 '16 at 2:26
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just one of those things, given $b_0=1, b_1 = 1,$ and $$ b_{n+2} = k \, b_{n+1} - b_n, $$ then $$ a_n = b_n^2. $$

The key step is $$ b_{n+2} b_n - b_{n+1}^2 = k-2 $$ Let us show how that fits in: we say $a_n = b_n^2.$ We know $b_{n+2} + b_n = k b_{n+1.}$ Square both sides, this gives $$ b_{n+2}^2 + a_n + 2 b_{n+2}b_n = k^2 a_{n+1}, $$ $$ b_{n+2}^2 + a_n = k^2 a_{n+1} - 2 b_{n+2}b_n, $$ $$ b_{n+2}^2 + a_n = k^2 a_{n+1} - 2 b_{n+2}b_n + 2 b_{n+2}b_n - 2 b_{n+1}^2 - 2 k + 4, $$ $$ b_{n+2}^2 + a_n = k^2 a_{n+1} - 2 b_{n+1}^2 - 2 k + 4, $$ $$ b_{n+2}^2 + a_n = k^2 a_{n+1} - 2 a_{n+1} - 2 k + 4, $$ $$ b_{n+2}^2 + a_n = (k^2 -2) a_{n+1} - 2 k + 4, $$ $$ b_{n+2}^2 = a_{n+2}. $$

In turn, $ b_{n+2} b_n - b_{n+1}^2 = k-2 $ comes from a standard quadratic forms construction. We consider the quadratic form $f(x,y) = x^2 - kxy + y^2.$ Its Hessian matrix is $$ H = \left( \begin{array}{rr} 2 & -k \\ -k & 2 \end{array} \right) $$

We define a matrix $A$ (for "automorphism") as $$ A = \left( \begin{array}{rr} k & -1 \\ 1 & 0 \end{array} \right). $$ The word automorphism means $$ A^T H A = H. $$ In turn, from the observation that $$ \left( \begin{array}{c} b_{n+2} \\ b_{n+1} \end{array} \right) \left( \begin{array}{rr} k & -1 \\ 1 & 0 \end{array} \right) = \left( \begin{array}{c} b_{n+1} \\ b_n \end{array} \right) $$ we find $$ b_{n+2}^2 - k b_{n+2} b_{n+1} + b_{n+1}^2 = b_{n+1}^2 - k b_{n+1} b_{n} + b_{n}^2, $$ so that $$ b_{n+1}^2 - k b_{n+1} b_{n} + b_{n}^2 $$ is independent of $n$ and constant. From $b_1$ and $b_0$ we find $$ b_{n+1}^2 - k b_{n+1} b_{n} + b_{n}^2 = 2-k. $$ Finally $$b_{n+2} b_n - b_{n+1}^2 = (k b_{n+1} - b_n) b_n - b_{n+1}^2 = - b_{n+1}^2 + k b_{n+1} b_n - b_n^2 = -(2-k) = k-2 $$

I asked the OP for the source of the problem, it turns out the part about defining $b_n$ was the hint given; exercise 24. The exercise (review?) set pdf is at PDF

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