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.
 A: For a cleverness free solution:
The number of positive roots counting multiplicity is $2$: the root $1$ is repeated twice.
You can use Descartes' rule of signs, which is a general method which can be useful sometimes.
Your equation (after multiplying by $x^2-1$) is
$$ P(x) = x^{2000} - 1000x^{1001} + 1000x^{999} - 1 = 0$$
which has $3$ sign changes.
So the number of positive roots is either $1$ or $3$ (counting multiple roots multiple times).
The derivative is $$P'(x) = 2000 x^{1999} - 1000\times 1001 x^{1000} + 1000 \times 999 x^{998}$$
Notice that $P(1) = 0$, $P'(1) = 0$.
Since $1$ is a root, and also of the derivative, the number of positive roots is $3$, counting $1$ at least two times.
But since we introduced an extraneous positive root by multiplying by $x^2 -1$, the number of positive roots of your original equation, counting multiplicity is $2$.
The second derivative
$$P''(x) = 2000\times 1999 x^{1998} - 1000\times1000\times1000 x^{999} + 1000 \times 999 \times 999$$
and $P''(1) = 0$. Thus the root $1$ is of multiplicity $2$.
This also applies to the equation $x^{2n} - n(x^{n+1} - x^{n-1}) - 1 = 0$ for $n \gt 1$.
See Also: Sturm's theorem.
A: 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 $1$.
A: I think a family of problems to consider would be $x^{2n} - n(x^{n + 1} - x^{n - 1}) - 1 = 0.$ Substituting $ n = 1000$ we get your case. 
