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I am taking a course in algebra and this problem was on my problem set, and I had no idea how to solve it.

Suppose we have a sequence $s_n$ of real numbers such that $5s_{n+1}-s_{n}-3s_{n}s_{n+1}=1$ for $1 \leq n \leq 42$ and $s_1=s_{43}$.

What are the possible values of $s_1+s_2+ \ldots + s_{42}$?

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  • $\begingroup$ The answer will be in terms of $s_1$, right? $\endgroup$ – SchrodingersCat Nov 16 '15 at 12:12
  • $\begingroup$ I'm not sure. I'd assume they would be in integer/real number form, but the question never specified and I was unable to get an answer. $\endgroup$ – user290610 Nov 16 '15 at 12:14
  • $\begingroup$ I think you might benefit from an explicit formula for $s_{n+1}$ via $s_n$. $\endgroup$ – Ivan Neretin Nov 16 '15 at 12:23
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Note that we have $$s_{n+1} (5-3s_n) = 1 + s_n,$$ thus in particular $s_n \neq 5/3$ for all $n$ and so $$\tag{1} s_{n+1} = \frac{1+s_n}{5-3s_n}.$$ The mapping $$f(z) = \frac{z + 1}{-3z+5}$$ is a Mobius transform. One can check that $$f^{(n)} (z) = \frac{az+ b}{cz+ d},$$ where $$\begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ -3 & 5\end{bmatrix}^n$$ Doing some linear algebra, we have $$\begin{bmatrix} 1 & 1 \\ -3 & 5\end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 3\end{bmatrix} \begin{bmatrix} 2 & 0 \\ 0 & 4\end{bmatrix}\begin{bmatrix} 1 & 1 \\ 1 & 3\end{bmatrix}^{-1}$$ When $n=42$, we have (direct computations) $$\begin{bmatrix} a & b \\ c & d \end{bmatrix} = \frac{1}{2} \begin{bmatrix} 3\cdot 2^{42} -4^{42} & -2^{42} + 4^{42} \\ 3\cdot 2^{42} - 3\cdot 4^{42} & 3\cdot 4^{42} - 2^{42} \end{bmatrix}$$ So the fact that $s_1 = s_{43}$ is the same as $$s_1 = \frac{(3\cdot 2^{42}-4^{42}) s_1 +(-2^{42} + 4^{42})}{(3\cdot 2^{42} - 3\cdot 4^{42})s_1 + 3\cdot 4^{42} - 2^{42}}$$ By some calculations, $$3s_1^2 - 4s_1 + 1 = 0\Rightarrow s_1= 1 \text{ or } \frac 13.$$ Put into $(1)$, we get either $s_n = 1$ for all $n$ or $s_n = \frac 13$ for all $n$. So $$s_1 + \cdots + s_{42} = 42 \text{ or } 14.$$

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Let $u_n = {3s_n-1 \over 1-s_n}$ (unless $s_n=1$ which gives us sum $42$). Then $u_{n+1} = {3{1+s_n \over 5-3s_n}-1 \over 1-{1+s_n \over 5-3s_n}} = {2 \over 4}{3s_n-1 \over 1-s_n} = {1 \over 2}u_n$; $u_{43}=u_1$ implies $u_1=0, s_1={1 \over 3}$.

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    $\begingroup$ While a nice demonstration, It is not immediately clear what would motivate one to define $u_n$ in this fashion, and thus to discover this useful property of it. Some explanation of what led you to this approach would be helpful. $\endgroup$ – Paul Sinclair Nov 16 '15 at 17:51
  • $\begingroup$ @PaulSinclair Well, there are several ways. I started from rewriting initial expression in factorized form (to rid of that annoying $s_ns_{n+1}$), and after several tries I got two forms : $(3s_{n+1}-1)(1-s_n)=2(s_n-s_{n+1})$ and $(3s_n-1)(1-s_{n+1})=4(s_n-s_{n+1})$, it was obvious from there. Essentially I used transformation corresponding to matrix $\pmatrix{1 && 1 \\ 1 && 3}^{-1}$ from John Ma's solution (but you have to know about Jordan form to get it that way). $\endgroup$ – Abstraction Nov 17 '15 at 8:58

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