# 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?

• Its x to the power 1000 , and x to the power 998 , and x to the power 999. i am sorry i tried but it was not coming in proper format. – vikiiii Mar 30 '12 at 3:26
• When exponents have multiple digits, you need to put curly braces (i.e., { } ) around them. – Matthew Conroy Mar 30 '12 at 3:30
• Since $(1-x^2)(1 + x^2 + x^4 + … + x^{998})= 1-x^{1000}$ we have $1 + x^2 + x^4 + … + x^{998} = \frac{1-x^{1000}}{1-x^2}$ so $$(x^{1000}+1)(1 + x^2 + x^4 + … + x^{998})=\frac{(1+x^{1000})(1-x^{1000})}{1-x^2}=\frac{1-x^{2000}}{1-x^2}$$ hence we are looking for solutions to $\frac{1-x^{2000}}{1-x^2}=1000x^{999}$. Equivalently, we want $$0=(1-x^{2000})-(1000x^{999})(1-x^{2000})=1000x^{2999}-x^{2000}-1000x^{999}+1.$$ – Alex Becker Mar 30 '12 at 3:52
• It should rather be $(1 - x^{2000} - 1000x^{999} + 1000x^{1001}) = 0.$ – Suresh Mar 30 '12 at 4:08

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 $n \gt 1$.

• @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...)". – Aryabhata Mar 30 '12 at 19:30
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 $1$.
• Just simply multiplying out $(1+x^{1000})(1 + x^2 + \dots + x^{998})$ gives us $1 + x^2 + x^4 + \dots + x^{998} + x^{1000} + \dots + x^{1998}$. Now we can apply AM $\ge$ GM. There is no need to bring $x^2 -1$ in. – Aryabhata Mar 30 '12 at 5:22
• There is no need to even multiply...You can apply the AM-GM inequality directly to $1+x^{1000}$ and $1 + x^2 + \dots + x^{998}$... – N. S. Mar 30 '12 at 5:36
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.