oVlad wrote: An arrangement of 12 real numbers in a row is called good if for any four consecutive numbers the arithmetic mean of the first and last numbers is equal to the product of the two middle numbers. How many good arrangements are there in which the first and last numbers are 1, and the second number is the same as the third? Let $\{x_n\}_{n=1}^{12}$ be a good arrangement. We have $x_{n+3}=2x_{n+2}x_{n+1}-x_n$ $\forall n\in\{1,2,...,9\}$ Adding the condition $x_1=1$ and $x_2=x_3=a$ : 1) If $|a|\le 1$ Setting $a=\cos t$, we quickly get $x_3=\cos 2t$ and $x_4=\cos 3t$ And an easy induction gives $x_n=\cos a_nt$ with $a_1=0$, $a_2=1$ and $a_{n+2}=a_{n+1}+a_n$ And so $x_n=\cos F_{n-1}t$ where $F_0=0,F_1=1$ and $F_{n+2}=F_{n+1}+F_n$ is the Fibonnacci sequence. So $x_{12}=\cos F_{11}t=\cos 89t$ And constraint $x_{12}=1$ is $89t=+2k\pi$ and $89$ such sequences. *** edited : erased the shift $\frac{\pi}2$ Thanks Tintarn for remark *** 2) If $a>1$ Setting $a=\cosh t$ with $t>0$, similar to above, we get $x_n=\cosh F_{n-1}t$ and so $\cosh 89t=1$ and no solution 3) If $a<-1$ Setting $a=-\cosh t$ with $t>0$, nearly similar to above, we get $x_n=\pm\cosh F_{n-1}t$ where $\pm$ is $+1$ if $n\equiv 1\pmod 3$ and $-1$ otherwise and so $|\cosh 89t=1|$ and no solution Hence the answer $\boxed{89\text{ such arrangements}}$

One can unify the approach slightly: By induction, $x_{12}=T_{89}(a)$ where $T_{89}$ is the $89$-th Chebyshev polynomial. But we have $89$ distinct solutions $a=\cos \frac{2k\pi}{89}$ for $k=0,1,2,\dots,88$. Since $T_{89}$ has degree $89$, there cannot be more than $89$ solutions. Done. (Btw, I believe the shift $+\frac{\pi}{2}$ in the above solution should not be there.)

Wait guys, I think there's only 45 arrangements. Since u can let the second and the third number be $x$ and then the last number is $T_{89} (x)=1$ so there are only 45 possible values for $x$.

Why $45$?

Tintarn wrote: Why $45$? We want to get the number of solutions of $T_{89} (x)=1$ and if you draw it in geogebra you will find 45 roots

Attachments:

Yes, now I understand. Indeed, since $T_{89}(x) \in [-1,1]$ for all $x \in [-1,1]$ (by its definition) all but one of the roots are double roots. Hence we get $1+\frac{89-1}{2}=45$ solutions. Thank you!