Scroll to see replies
rs=1, \text{ so } s = -1/r \text { and so } r^2+3/r^2 = 4 \implies r^4-4r^2+3 = 0\\[br]\implies r^2 = 1, 3
2z=2\pm\sqrt{4 - 4(1+i)(2\sqrt{3}-2)} \\[br]\implies z = 1 \pm \sqrt{1 - (1+i)(2\sqrt{3}-2)} = 1 \pm \sqrt{3-2\sqrt{3}+2(1-\sqrt{3})i)
F_{k+2}F_k - F_{k+1}^2 \\[br]= (F_{k+1}+F_k)F_k - F_{k+1}^2 \\ [br]= F_k^2 +F_{k+1}F_k - F_{k+1}^2 = F_k^2 - F_{k+1}(F_{k+1} - F_k) \\[br]= [F_{k+1}F_{k-1}+(-1)^{k+1}] - F_{k+1}(F_{k+1} - F_k) \\[br]= (-1)^{k+1} + F_{k+1}F_{k-1} - F_{k+1}(F_{k-1})\\[br]= (-1)^{k+1}F_{k+1}^2
F_{n+k+1} = F_{n+k}+F_{n+k-1}\\[br]= F_kF_{n+1}+F_{k-1}F_n + F_k-1F_{n+1}+F_{k-2}F_n\\[br]=(F_k+F_{k-1})F_{n+1}+(F_{k-1}+F_{k-2}F_n \\[br]=F_{k+1}F_{n+1}+F_k F_n
pr=qs \iff|y+1||y-1| = |x-1||x+1|\\[br]\iff (y+1)^2(y-1)^2 = (x-1)^2(x+1)^2 \\[br]\iff [(y+1)(y-1)]^2 - [(x+1)(x-1)]^2 = 0 \\[br]\iff (y^2-1)^2 - (x^2 - 1)^2 = 0 \\[br]\iff y^4-2y^2 - x^4 + 2x^2 = 0\\[br]\iff y^4-x^4 -2(y^2-x^2) = 0\\[br]\iff (y^2-x^2)[(y^2+x^2)-2] = 0\\[br]\iff (y+x)(y-x)[(y^2+x^2)-2] = 0
pr=qs \iff|y+1||y-1| = |x-1||x+1|\\[br]\iff (y+1)^2(y-1)^2 = (x-1)^2(x+1)^2 \\[br]\iff [(y+1)(y-1)]^2 - [(x+1)(x-1)]^2 = 0 \\[br]\iff (y^2-1)^2 - (x^2 - 1)^2 = 0 \\[br]\iff y^4-2y^2 - x^4 + 2x^2 = 0\\[br]\iff y^4-x^4 -2(y^2-x^2) = 0\\[br]\iff (y^2-x^2)[(y^2+x^2)-2] = 0\\[br]\iff (y+x)(y-x)[(y^2+x^2)-2] = 0