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.

Thanks in advance.

  • $\begingroup$ Do you have values for the variable $c$ ? $\endgroup$ Mar 21 '19 at 18:53
  • $\begingroup$ The solution of Maple is a polynomial of degree 64! $\endgroup$ Mar 21 '19 at 18:55
  • $\begingroup$ hey Dr! both N and c are positive integers! So does that mean this is unsolvable? $\endgroup$
    – PetaGlz
    Mar 21 '19 at 18:58
  • $\begingroup$ It is solvable, but i think no per hand! $\endgroup$ Mar 21 '19 at 18:59
  • $\begingroup$ @Dr.SonnhardGraubner ok...so could u show me some possible approaches? $\endgroup$
    – PetaGlz
    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 $t > 0$.

  • 1
    $\begingroup$ That is very clear and fast. Thanks a lot! $\endgroup$
    – PetaGlz
    Mar 21 '19 at 19:32

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