$x^{x^2} + x^{x^8} =?$ Given $x^{x^4} = 4$ If x is any complex number such that  $x^{x^4} = 4$ , then find all the possible values of :
$x^{x^2} + x^{x^8}$
First, I used laws of exponents to give $18$ as answer. However , I realised that I've misused it. Further I used logarithms which yielded
$$ x^{x^2} + x^{x^8} = 4^{1/x^2} + 4^{x^2} $$
After this, I am stuck and can't proceed further. By hit and trial, $x = \sqrt{2}$ seems to be one of the solution. Any help is appreciated.

The value of $x^{x^2}+x^{x^8}$ is not uniquely determined by the condition $x^{x^4}=4$ if we work in the complex numbers, as can be observed by considering $x=\sqrt 2$ and $x=i\sqrt 2$. So I'd say that the problem is ill-posed. The original problem is probably intended to be posed in the context of real numbers.
