The condition $ \alpha>0 $ seems to be unnecessary.

Of course it is; from the relation $\alpha x_1 = x_1 + x_2$ for $n=1$, follows $x_2=\alpha - 1$, so we need $\alpha > 1$. It is probably given just from routine, to completely take negative numbers out of the picture.

I found $\alpha$ has to be at least four. Define $t_n=x_1+x_2+\cdots+x_n$ and $t_{n+2}=\alpha(t_{n+1}-t_n)$. The sequence $(t_n)_{n\geq 1}$ is increasing row when $\alpha$ is at least four. Answer: 4 as it works.

This problem similar to http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1353587&sid=03f3265a58d7cc6abeecd8709d9f1cb9#p1353587