jmc

We can factor a constant out of the first radical: $$\begin{array}{rl}\sqrt{4+\sqrt{8+4c}}& =\sqrt{4+\sqrt{4(2+c)}}\\ & =\sqrt{4+2\sqrt{2+c}}\\ & =\sqrt{2(2+\sqrt{2+c})}\\ & =\sqrt{2}\sqrt{2+\sqrt{2+c}}.\end{array}$$Then, we can combine like terms and solve: $$\begin{array}{rl}\sqrt{2}\sqrt{2+\sqrt{2+c}}+\sqrt{2+\sqrt{2+c}}& =2+2\sqrt{2}\\ \Rightarrow (1+\sqrt{2})\sqrt{2+\sqrt{2+c}}& =2(1+\sqrt{2})\\ \Rightarrow \sqrt{2+\sqrt{2+c}}& =2\\ \Rightarrow 2+\sqrt{2+c}& =4\\ \Rightarrow \sqrt{2+c}& =2\\ \Rightarrow 2+c& =4\\ \Rightarrow c& =\boxed{2}\end{array}$$