About the solvability of certain equations How one can see if this equation has real solutions:

$$x^{2^{k}}-x-a=0$$ where $x$ is the unknown, $k$ is a positive integer and $a$ is a real constant.
Use Descartes' Rule of Sign: Looking at the polynomial

$$ 1\cdot x^{2^k} +0\cdot x^{2^k-1} + 0\cdot x^2+1\cdot x -a =0$$

you can count the number of sign changes in the sequence of coefficients. The number of sign changes is an upper bound on the number of positive roots: If there are $c$ sign changes, then the polynomial has $c-2k$ positive roots, where $k$ is integer. Hence, if $c=1$, the polynomial has exactly one positive root, if $c=2$, the polynomial has either no or two positive roots, etc.

In our case the sequence of coefficients is $(1,0,0,\dots,0,1,-a)$. There is exactly one sign change in the polynomial, hence there is exactly one positive root. If you replace $x$ by $-x$, you get

$$ x^{2^k} +x -a =0$$

which has again only one sign change. Hence there is exactly one negative root.
Use Descartes' Rule of Sign: Looking at the polynomial

$$ 1\cdot x^{2^k} +0\cdot x^{2^k-1} + 0\cdot x^2+1\cdot x -a =0$$

you can count the number of sign changes in the sequence of coefficients. The number of sign changes is an upper bound on the number of positive roots: If there are $c$ sign changes, then the polynomial has $c-2k$ positive roots, where $k$ is integer. Hence, if $c=1$, the polynomial has exactly one positive root, if $c=2$, the polynomial has either no or two positive roots, etc.

In our case the sequence of coefficients is $(1,0,0,\dots,0,1,-a)$. There is exactly one sign change in the polynomial, hence there is exactly one positive root. If you replace $x$ by $-x$, you get

$$ x^{2^k} +x -a =0$$

which has again only one sign change. Hence there is exactly one negative root.