P.S. edited

I edited the problem to another one which I guess is right. This is supposed to be ILL 1969. If anybody can find the exact statement, please say it.

Incomplete solution Let $p,q $ is roots of equation $x^2-\alpha x- \beta =0$. Let this system has solution in $a, b$: $(pa+qb)=x_0$; $p^2a+q^2b=x_1$ (1). Let prove, that $x_i=p^{i+1}a+q^{i+1}b$ satisfied to condition. We have: $0=a(p^{k+2}-\alpha p^{k+1}- \beta p^{k})$ ($k \geq 1$) and $0=b(q^{k+2}-\alpha q^{k+1}- \beta q^{k})$ ($k \geq 1$), adding this two equation we have $x_{k+1}-\alpha x_k- \beta x_{k-1}=0$=> $x_i=p^{i+1}a+q^{i+1}b$ work. Now if system (1) haven't solution in $a,b$. But then holds one of this: 1)$p=0$;$q=0$, 2)$p=0$;$q \neq 0$ 3)$p=q$, $x_1 \neq x_0 p$. 1)<=> $\alpha = \beta =0$, obviously $x_k=0$ for $k \geq 2$; 2)<=> $\alpha \neq \beta =0$. then $x_k=\alpha^{k-1}x_1$; 3)<=>$ \boxed{\alpha = \pm 2\sqrt{-\beta}, x_1 \neq px_0}$ I don't know solution to this case

But this is the theory of linear recurrence relations. The characteristic equation is $x^2 - \alpha x - \beta = 0$. Its discriminant is $\Delta = \alpha^2 + 4\beta$. For $\Delta \neq 0$ there are two distinct roots $\rho_1$, $\rho_2$ (real or complex, according to the sign of $\Delta$). The general solution is $x_n = p\rho_1^n + q\rho_2^n$, where the coefficients $p$ and $q$ are determined from the system for values $n=0$ and $n=1$. For $\Delta = 0$ there is a double root $\rho$. The general solution is $x_n = (pn + q)\rho^n$, where the coefficients $p$ and $q$ are determined from the system for values $n=0$ and $n=1$.