# Solving algebra with multiple square root

I am currently solving an algebra and can't figure it out, could anyone help me on this?

$$2\sqrt{N + \sqrt{N^2+4c^2}} = \sqrt{N + \sqrt{N^2+3c^2}} + \sqrt{N + \sqrt{N^2+5c^2}}$$

Which I would like to have a solution to represent N in terms of c, or c in terms of N, either way works. If this is unsolvable, please show me the reason for that.

• Do you have values for the variable $c$ ? Mar 21 '19 at 18:53
• The solution of Maple is a polynomial of degree 64! Mar 21 '19 at 18:55
• hey Dr! both N and c are positive integers! So does that mean this is unsolvable? Mar 21 '19 at 18:58
• It is solvable, but i think no per hand! Mar 21 '19 at 18:59
• @Dr.SonnhardGraubner ok...so could u show me some possible approaches? Mar 21 '19 at 19:00

Of course $$c=0$$ is a solution. But there are no other real solutions.
After noting that $$N=0$$ is impossible, we divide both sides by $$\sqrt{N}$$ and let $$c^2/N^2 = t$$ to get $$2 \sqrt{1+\sqrt{1 + 4 t}} = \sqrt{1+\sqrt{1+3t}} + \sqrt{1+\sqrt{1+5t}}$$ We can write this as $$2 g(4 t) = g(3 t) + g(5 t)$$ where $$g(x) = \sqrt{1+\sqrt{1+x}}$$ Now this function is strictly concave for $$x > 0$$, as we see by taking its second derivative. Thus since $$4 = (3+5)/2$$, $$g(4t) > (g(3t) + g(5t))/2$$ for $$t > 0$$.