Define a sequence {$a_n$}$^{\infty}_{n=1}$ by $a_1 = 4, a_2 = a_3 = (a^2 - 2)^2$ and $a_n = a_{n-1}.a_{n-2} - 2(a_{n-1} + a_{n-2}) - a_{n-3} + 8, n \ge 4$, where $a > 2$ is a natural number. Prove that for all $n$ the number $2 + \sqrt{a_n}$ is a perfect square.
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Tags: algebra, polynomial, inequalities, algebra unsolved
123steeve
19.12.2011 18:19
The first few terms of the sequence are \[a_1=4,a_2=x,a_3=x,a_4=(x-2)^2,a_5=x(x-3)^2,...\] where $x=(a^2-2)^2$. Put $f_n(x)=a_n$ and let $b_i=a_i-2$ so that $b_1=2,b_2=x-2,b_3=x-2,...$ with the defining equation now $b_n=b_{n-1}b_{n-2}-b_{n-3}$ and $b_n+b_{n-3}=b_{n-1} b_{n-2}$ We prove two results by induction. (1) $P(n)=b_n^2+b_{n-1}^2+b_{n-2}^2-b_n b_{n-1}b_{n-2}-4=0$ Base result for $P(3)$ clearly true. Assume true $\forall n \le k-1$ \[P(k)-P(k-1)=b_k^2-b_{k-1}b_{k-2}[b_k-b_{k-3}]-b_{k-3}^2=\]\[=b_k^2-[b_k+b_{k-3}][b_k-b_{k-3}]-b_{k-3}^2=0\]as required (2) $f_n(x)=f_n[(a^2-2)^2]=[f_n(a^2)-2]^2.$ Clearly true for $n=1,2,3$ Assume true $\forall n \le k-1$ \[f_n(x)=[f_{n-1}(x)-2][f_{n-2}(x)-2]-[f_{n-3}(x)-2]+2\]i.e.\[f_n[(a^2-2)^2]=[p^2-2][q^2-2]-[r^2-2]+2\] by hypothesis, where $p=f_{n-1}(a^2)-2,...$ etc. Then, using the first result,\[f_n[(a^2-2)^2]=p^2q^2-2[p^2+q^2]-r^2+8=\]\[=p^2q^2-2[4-r^2+pqr]-r^2+8=(pq-r)^2=[f_n(a^2)-2]^2\] as required. Hence $2+\sqrt{a_n}=f_n(a^2) \in \mathbb{N}$ as required. Can I ask where you got this question? Using WolframAlpha, I found up to $a_8=x(x^6-13x^5+65x^4-156x^3+182x^2-91x+13)^2$ and finding the maximums give very nice graphs.
spanferkel
24.12.2011 01:43
123steeve wrote: Using WolframAlpha, I found up to $a_8=x(x^6-13x^5+65x^4-156x^3+182x^2-91x+13)^2$ and finding the maximums give very nice graphs. Yes indeed, $a_8=2T_{13}(\frac x2-1)+2$ and generally for $n\ge 0$ we have $\frac{a_{n+2}}2-1=T_{F_n}(\frac x2-1)$ where $T_k$ are the Chebyshev polynomials. and $F_k$ the Fibonacci numbers ($F_0=F_1=1$). For proving this, the recursion formula $T_{n+m}=2T_nT_m-T_{n-m}$ (for $n\ge m$) comes in handy.