Two sequences, So weird !

That recurrence relation sure looks suspiciously familiar... Familiar to how to rotate a rectangular coordinate! This makes us think of using polar coordinates.

Let $\alpha=r\cos\theta$ and $\beta = r\sin \theta$

We have that $\left\{\begin{array}{l} a_{n+1}=(r\cos\theta)a_n-(r\sin\theta)b_n\\ b_{n+1}=(r\sin\theta)a_n+(r\cos\theta)b_n\end{array}\right.$

This is exactly the rotation formula for rectangular coordinates. Putting the coordinates on the Argand Diagram, we see that the recurrence relations simply turn into $a_{n+1}+ib_{n+1}=r\text{cis}\theta(a_n+ib_n)$

Letting $a_n+ib_n=z_n$ , we see $z_{n+1}=r\text{cis}\theta(z_n)$

Then, since $z_1=r\text{cis}\theta$ we have

$z_n=(r\text{cis}\theta)^n=z_1^n$

Note that because $a_{1997}=b_1$ and $b_{1997}=a_1$ , we must have $||z_{1997}||=||z_1||$ . Thus, $r=1$ .

Now we just need to find a way to represent $a_{1997}=b_1$ and $b_{1997}=a_1$ .

Note that $z_1^{1997}=a_{1997}+ib_{1997}=b_1+ia_1=i\overline{z_1}$

Where the middle inequality makes sure that $a_{1997}=b_1$ and $b_{1997}=a_1$ is satisfied.

Thus, we want to solve $z_1^{1997}=i\overline{z_1}$

Multiplying both sides by $z_1$ gives $z_1^{1998}=i||z_1||$

But $||z_1||=1$ since $r=1$ . Thus, $z_1^{1998}=i$

This is a polynomial with degree $1998$ ; thus, there must be $\boxed{1998}$ solutions.

This is from the 1997 Turkish MO. I'm one to talk, but please at least give credit. If you're going to take a problem from somewhere else, at least change it a little bit. :O