F[n] = 1/sqrt(5) * { [ (1 + sqrt(5)) / 2 ]^n - [ (1 - sqrt(5)) / 2 ] ^ n } \$s = sqrt(5) Therefore: [ Given limits according to the |q| <= 1 ] A_F(x) = 1/\$s * { x(1+\$s)/2/[ 1 - x(1+\$s)/2] - x(1-\$s)/2/[1 - x(1-\$s)/2] } = x/(2*\$s) * { (1+\$s)/[1 - x(1+\$s)/2] - (1-\$s)/[1 - x(1-\$s)/2] } = x/(2*\$s) * { 1/[1/(1+\$s) - x/2] - 1/[1/(1-\$s) - x/2] } = [y = x/2] y*\$s * {1/[1/(1+\$s) - y] - 1/[1/(1-\$s) - y] } = [z = 1/y] \$s * { 1 / [z/(1+\$s) - 1] - 1 / [z/(1-\$s) - 1] } = N \$s * [z/(1-\$s)-1 - z/(1+\$s) + 1 ]/[z/(1+s)-1]/[z/(1-s)-1] = N \$s * z * (1/(1-\$s)-1/(1+\$s)) = N[z/(1+\$s)-1][z/(1-s)-1] [ * (1+\$s) * (1-\$s)] \$s * \$z * (1+\$s-1+\$s) = N[z-(1+\$s)][z-(1-\$s)] 2 \$s^2 * \$z = N [ z^2 - 2z + (1-\$s^2)] \$s^2 = 5 10z = N [ z^2 - 2z - 4 ] z^2 - (2+10/N)z - 4 = 0 z = [ -b +/- sqrt(b^2-4ac) ] / (2*a) z is rational iff b^2-4ac is a square of a rational number: ∆ = (2+10/N)^2+16 = 4[ (1+5/N)^2+4 ] = 4[ (N+5)^2/N^2 + 4 ] ∆ = 4[ [ (N+5)^2+4N^2 ] / N^2 ] rational iff (N+5)^2+4N^2 is a whole square.