TITLE: Prove that the equation $x^{10000} + x^{100} - 1 = 0$ has a solution with $0 < x < 1$
QUESTION [18 upvotes]: Prove that the equation $x^{10000} + x^{100} - 1 = 0$ has a solution with $0 < x < 1$.

This is a homework question. I know I could probably find a solution that would complete the proof, but I don't think that is what this question is asking. What proof-techniques should I use to prove this is true?

REPLY [6 votes]: I'm trying to improve Qiaochu Yuan's bounds.
The equation
 $$x^{100}+x^{10\thinspace000}=1 \qquad\qquad(1)$$
obviously has exactly one solution $\xi\in\ ]0,1[\ $. We put $100=:n$ and
 $$x:=1-{y\over n^2}$$
 with a new unknown $y>0$. In this way equation $(1)$ becomes
 $$\Bigl(f(y):=\Bigr)\quad1-\Bigl(1-{y\over n^2}\Bigr)^n=\Bigl(1-{y\over n^2}\Bigr)^{n^2}\quad\Bigl(=: g(y)\Bigr)\ .\qquad\qquad(2)$$
 In order to get some indication on the order of magnitude to be expected we assume $y\ll n$ and then have approximatively $$f(y)\doteq {y\over n} \>, \quad g(y)\doteq e^{-y}\ .$$ In this way equation  $(2)$ morphs into the simple equation $y\, e^y=100$ with the solution $y_*\doteq 3.3856$. It will turn out that this value is a very good approximation to the true solution $\eta$ of $(2)$.
The function $f$ is monotonically increasing for $0<y<n^2$, and the function $g$ is monotonically decreasing there. In the following we shall consider the two $y$-values $y_1:=3.38$ and $y_2:=3.40$, and (with the help of a pocket calculator) we shall prove that
 $$f(y_1)<g(y_1)\>, \quad f(y_2)> g(y_2)\ .\qquad\qquad(3)$$
This implies $y_1<\eta<y_2$, resp. $$1-0.000340<\xi<1-0.000338\ .$$ 
So we need bounds for $f$ and $g$. Concerning $f$ it is easily seen that
 $${y\over n}\Bigl(1-{y\over 2n}\Bigr)<f(y)<{y\over n}\qquad (0<y< n^2)$$
  – the upper bound being nothing else but Bernoulli's inequality. This immediately implies
 $$f(3.38)<0.0338\>,\qquad f(3.40)>0.0340\,(1-0.0170)=0.03342\ .$$
 For $g$ we begin with
 $$\log(1-t)=-\Bigl(t+{t^2\over2}+{t^3\over3}+\ldots\Bigr)\  \cases{\  <-t \cr \  >-t -{\displaystyle{t^2/2\over 1-t}} \cr}\qquad (0<t<1)\ .$$
 Putting $t:={\displaystyle{y\over n^2}}$ and multiplying with $n^2$ we get the estimates
 $$-y -{y^2 \over 2(n^2-y)}< \log \bigl(g(y)\bigr)< -y\ ,$$
 and using the inequality ${\displaystyle{\exp\Bigl({-y^2\over 2(n^2-y)}\Bigr)>1-{y^2\over 2(n^2-y)}}}$ we therefore obtain
 $$e^{-y}\Bigl(1-{y^2\over 2(n^2-y)}\Bigr)< g(y)< e^{-y}\ .$$
This implies
 $$g(3.38)> 0.034047\,(1-0.000571)=0.034028\>,\quad g(3.40)<0.03337\ .$$
 It follows that the values of $f$ and $g$ at the places $y_1$ and $y_2$ obey the stated relation $(3)$.