help me prove this

$\textbf{Here $\mathbf{f(x+a)=\frac{1}{2}+\sqrt{f(x)-(f(x))^2}}$}$\\\\\\ \mathbf{\left(f(x+a)-\frac{1}{2}\right)^2=f(x)-(f(x))^2............................(1)}$\\\\\\ \textbf{Now $\mathbf{x\rightarrow x+a}$,We Get}$\\\\ \mathbf{\left(f(x+2a)-\frac{1}{2}\right)^2=f(x+a)-(f(x+a))^2............................(2)}$\\\\\\ $\mathbf{\left(f(x+2a)-\frac{1}{2}\right)^2=\frac{1}{2}+\sqrt{f(x)-(f(x))^2}-\frac{1}{4}-f(x)+(f(x))^2-\sqrt{f(x)-(f(x))^2}}\\\\\\ \mathbf{\left(f(x+2a)-\frac{1}{2}\right)^2=\left(f(x)-\frac{1}{2}\right)^2}$\\\\\\ \textbf{So $\mathbf{f(x+2a)-\frac{1}{2}=\pm \left(f(x)-\frac{1}{2}\right)}}$\\\\\\ \textbf{\boxed{\boxed{\mathbf{f(x+2a) = f(x)}$}} and $\mathbf{f(x+2a)+f(x)-1=0}$}

thanks a lot sir

Okay I didnt notice there was condition on f(x).