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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.

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    It should rather be $(1 - x^{2000} - 1000x^{999} + 1000x^{1001}) = 0.$2012-03-30

3 Answers 3

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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.

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    @vikiiii: Tao's answer has it: "since $1 + x^2 + \dots + x^{2m-2} \ge m x^{m-1}$ (algeraic average is great than blah...)".2012-03-30
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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 $1$.

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    There is no need to even multiply...You can apply the AM-GM inequality directly to $1+x^{1000}$ and $1 + x^2 + \dots + x^{998}$...2012-03-30
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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.

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    This is more suited to be a comment.2012-03-30