mestavk wrote: $a_{n+1} = \alpha a_{n} - \beta b_{n}$, $\beta_{n+1} = \beta a_{n} + \alpha b_{n}$ Do you mean $a_{n+1} = \alpha a_{n} - \beta b_{n}$, $b_{n+1} = \beta a_{n} + \alpha b_{n}$ [Moderator: Yeah he did, it has now been fixed.]

You can easily get that $a^2+b^2$ will be constant iff $\alpha^2+\beta^2=1$ and otherwise it can't. Then we can substitute $cos{x},sin{x}$ and it seems it will work (not done completely *** so yes, just a typo

If that, we can write $ a^2_n+b^2_n = (\alpha a_{n-1}-\beta b_{n-1})^2+(\beta a_{n-1}+\alpha b_{n-1})^2=(\beta^2 +\alpha^2)(a^2_{n-1}+b^2_{n-1})=...=(\beta^2 +\alpha^2)^{n} $ From the hypothesis we have $a_{1997}= b_{1}, b_{1997}= a_{1} \implies \beta^2 +\alpha^2=a^2_{1997}+b^2_{1997}=(\beta^2 +\alpha^2)^{1997} $ Then $ \beta^2 +\alpha^2 = 1 $ Note that the case $ \beta^2 +\alpha^2 = 0 $ should be neglected.

Note that \[ a_{n+1}^2 + b_{n+1}^2 = \bigl(\alpha^2+\beta^2\bigr)\bigl(a_n^2+b_n\bigr), \]hence $\alpha^2+\beta^2=1$. Reparameterizing $(\alpha,\beta)=(\cos\theta,\sin\theta)$ for $0\le \theta\le \pi/2$, and then casting the system in matrix form, we arrive at \[ \begin{pmatrix} a_{n+1}\\ b_{n+1} \end{pmatrix} =\underbrace{\begin{pmatrix} \cos\theta & -\sin\theta \\\sin\theta &\cos\theta \end{pmatrix}}_{\triangleq R_\theta}\begin{pmatrix} a_{n}\\ b_{n} \end{pmatrix}. \]Clearly, $R_\theta$ is a rotation, and it is easily seen that $R_{\theta_1}R_{\theta_2}=R_{\theta_1+\theta_2}$ for every $\theta_1,\theta_2\in\mathbb{R}$. Having done the math, we arrive at $(a_{1997},b_{1997})=(b_1,a_1)$ iff \[ \cos(1997\theta)=\sin(\theta)\qquad\text{and}\qquad \sin(1997\theta)=\cos(\theta). \]The rest is easy.

Let's define $S_n = a_{n}^2+b_{n}^2$. It is given that $S_1=S_{1997}=\alpha ^2+ \beta ^2$. Let's square the general term equations: $$ \begin{array}{rcl} a_{n+1}^2 &=& \alpha^2 a_{n}^2+\beta^2 b_{n}^2 - 2 \alpha \beta a_n b_n \\ b_{n+1}^2 &=& \beta^2 a_{n}^2+\alpha^2 b_{n}^2 + 2 \alpha \beta a_n b_n\\ a_{n+1}^2 + b_{n+1}^2 &=& (\alpha ^2+\beta^2)(a_n^2+b_n^2)\\ S_{n+1} &=& (\alpha ^2+\beta^2)S_n \end{array} $$ The sequence $S_n$ will be decreasing for $0<\alpha ^2+\beta^2<1$ and increasing for $1<\alpha ^2+\beta^2$. Since $S_1 \neq S_{1997}$ in both cases, there is no solution. For $\alpha ^2+\beta^2=0$, the solution is $\alpha = \beta = 0$ ($a_n=b_n=0$). For $\alpha ^2+\beta ^2=1$, let's assume $\alpha = \cos x$ and $\beta = \sin x$. Then, $a_2 = \alpha^2 - \beta^2 = \cos^2x - \sin^2 x = \cos 2x$ and $b_2=2\alpha \beta = 2\cos x\sin x = \sin 2x$. Continuing this pattern, we get $a_n = \cos nx$ and $b_n=\sin nx$. Alternatively, by induction, we can verify $a_{n+1}= \cos (n+1)x$ and $b_{n+1}= \sin (n+1)x$. Now, we have $a_{1997}=b_1$. This leads to the equations: $$ \begin{array}{lcl} \cos 1997x &=& \sin x = \cos (\pi/2 - x) = \cos (3\pi /2+x) \\ \sin 1997x &=& \cos x = \sin (\pi/2 - x) = \sin (\pi /2+x) \end{array} $$ For the first equation (a), two cases arise: 1. $1997x = \pi/2-x+ 2k\pi \Rightarrow 1998x = \frac{(4k+1)\pi}{2} \Rightarrow x = \frac{(4k+1)\pi}{3996}$ 2. $1997x = 3\pi/2+x+ 2k\pi \Rightarrow 1996x = \frac{(4k+3)\pi}{2} \Rightarrow x = \frac{(4k+3)\pi}{3992}$ For the second equation (b), the cases are: 1. $1997x = \pi/2-x+ 2k\pi \Rightarrow 1998x = \frac{(4k+1)\pi}{2} \Rightarrow x = \frac{(4k+1)\pi}{3996}$ 2. $1997x = \pi/2+x+ 2k\pi \Rightarrow 1996x = \frac{(4k+1)\pi}{2} \Rightarrow x = \frac{(4k+1)\pi}{3992}$ There is no common solution for the cases in (a.2) and (b.2). For the cases in (a.1) and (b.1), the common solution is $x = \frac{(4k+1)\pi}{3996}$. Therefore, the solutions are $(\alpha, \beta) \in \{ (\cos \frac {\pi}{3996}, \sin \frac {\pi}{3996}), (\cos \frac {5\pi}{3996}, \sin \frac {5\pi}{3996}), \dots , (\cos \frac {(4\cdot 1997 + 1)\pi}{3996}, \sin \frac {(4\cdot 1997+1)\pi}{3996})\}$, and this set has exactly 1998 elements, as $4 \nmid 4k_1+4k_2+2$ for any two elements $(\cos \frac{(4k_1+1)\pi}{3996}, \sin \frac{(4k_1+1)\pi}{3996} )$ and $(\cos \frac{(4k_2+1)\pi}{3996}, \sin \frac{(4k_2+1)\pi}{3996} )$. So the total number of solutions is equal to $1+1998=1999$. Source: Mathematical Olympiads 1997-1998 Problems and Solutions From Around The World Note: The book "Mathematical Olympiads 1997-1998 Problems and Solutions From Around The World" gives the answer as 1998. The pairs that satisfy this are given as $(0, 0)$ and for $k=1,3,\dots , 3997$, the pairs $(\cos \frac {\pi}{3998}k, \sin \frac {\pi}{3998}k)$. Therefore, if someone could review the solution I provided, I would appreciate it.