Periodicity of a Function Given the Functional Equation $f(x+a)=\frac12+\sqrt{f(x)-\big(f(x)\big)^2}$ [closed]
$$f(x+a)\big(1-f(x+a)\big)\\ =\left(\frac{1}{2}+\sqrt{f(x)\big(1-f(x)\big)}\right)\left(\frac{1}{2}-\sqrt{f(x)\big(1-f(x)\big)}\right)\\ =\frac{1}{4}-f(x)\big(1-f(x)\big)$$ $$\therefore f(x)\big(1-f(x)\big)+f(x+a)\big(1-f(x+a)\big)=\frac{1}{4}$$ $$\therefore f(x+a)\big(1-f(x+a)\big)+f(x+2a)\big(1-f(x+2a)\big)=\frac{1}{4}$$ Thus subtracting the above two equations $$\therefore f(x)\big(1-f(x)\big)=f(x+2a)\big(1-f(x+2a)\big)$$ $$\therefore f(x+3a)=\frac{1}{2}+\sqrt{f(x+2a)\big(1-f(x+2a)\big)}\\ =\frac{1}{2}+\sqrt{f(x)\big(1-f(x)\big)}=f(x+a)$$ $$\therefore f(x+2a)=f(x)$$