<s>[INST] Number of positive solutions? What is the number of positive solutions to 
$$ (x^{1000} + 1)(1 + x^2 + x^4 + \cdots + x^{998}) = 1000x^{999}? $$ 
I tried to solve it.
First I used by using sum of Geometric Progression.
Then the equation becomes too complicated and is in the power of 1998.
How can I get the number of positive solutions with that equation?
Thanks in advance.
 [/INST] Consider$$(x^{m} + 1)(1 + x^2 + x^4 + \cdots + x^{m-2}) = mx^{m-1}?$$
by multiplying $x^2-1$ in tow sides of equation,we have
$$(x^{m} + 1)(x^m-1)=mx^{m-1}(x^2-1)$$,
Last we have
$$x^{2m} - m(x^{m + 1} - x^{m - 1}) - 1 = 0 \tag{#}$$
Factorization
$$\begin{align*}
&x^{2m} - m(x^{m + 1} - x^{m - 1}) - 1\\
&=(x^{2m}-1)-mx^{m-1}(x^2-1)\\
&=(x^2-1)(1+x^2+x^4+\cdots x^{2m-2})-mx^{m-1}(x^2-1)\\
&=(x^2-1)(1+x^2+x^4+\cdots x^{2m-2}-mx^{m-1})\\
\end{align*}$$
since $1+x^2+x^4+\cdots x^{2m-2}\geq mx^{m-1}$(algebraic average is great than geometry average $\forall x>0$)if and only if $1=x^2= \cdots$ ,then$$1+x^2+x^4+\cdots x^{2m-2}= mx^{m-1}$$
so the root of equation(#)is $1$ and -$1$,but sine we first multiply with$(x^2-1)$,
so maybe there are some  extraneous roots in it,substitute $1$ and$-1$ for original equation,we conclude that the root of original equation is $1$.
</s>