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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.

  • 0
    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.2012-03-30
  • 0
    When exponents have multiple digits, you need to put curly braces (i.e., { } ) around them.2012-03-30
  • 0
    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 $$0=(1-x^{2000})-(1000x^{999})(1-x^{2000})=1000x^{2999}-x^{2000}-1000x^{999}+1.$$2012-03-30
  • 1
    It should rather be $(1 - x^{2000} - 1000x^{999} + 1000x^{1001}) = 0.$2012-03-30

3 Answers 3