(x_(n+1), y_(n+1))=(x_n^2-y_n^2+a, 2x_ny_n +b+2). (i) Find (x1, y1) if (a, b)=(1,-1) and (x_n, y_n) is constant. (ii) Find (a, b) if (x1, y1)=(-1,1) and (x_n, y_n) has period 2.

(i) x_1=x_1^2-y_1^2+1, y_1=2x_1y_1+1 by the equations given and the equality of x_1, x_2, x_3_. Substituting and trial and error of factor theorem results in x_1(x_1-1)(4x_1^2-4x_1+5)=0 . The quadratic has no real roots so the pairs of solutions are (x_1, y_1)=(0, 1), (1, -1). (ii) (x_2, y_2)=(a,b), (x_3,y_3)=(a^2-b^2+a, 2ab+b+2)=(x_1, y_1) by its periodic nature. Subbing a from the y equation into the x equation gets a quadratic in b^2, which has one +ve and one -ve solution. Since b is real, b>0, implying b^2=4/3. This results in the answers being (a, b)=(-1/2-rt3/4, 2/rt3), (-1/2+rt3/4, -2/rt3).