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The number of natural number $n$ in the interval $[1005,2010]$ for which the polynomial

$$1+x+x^2+x^3\dots +x^{(n-1)}$$ divides the polynomial $$1+x^2+x^4\dots+x^{2010}$$ is:

I could realize that the terms in the two polynomials are in GP and I applied the sum to n terms of a GP formula and then divided both the polynomials.

What I get is $(x^{2010}-1)/(x^{n}-1)$.

Now how to proceed?

Andrés E. Caicedo
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Rajath Radhakrishnan
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  • Your second formula is a sum over $x^{2n}$ instead of $x^n$, how does that affect this scenario? Also, isn't the final formula with $2011$ and $n+1$? – abiessu Aug 24 '13 at 14:09
  • yeah got the mistake but my ans comes out to be (x^(2012)-1)/(x+1)(x^(n)-1)......but not able to proceed further..now how to proceed? – Rajath Radhakrishnan Aug 24 '13 at 14:28
  • @RajathKrishnaR I made lots of edits but still need your refinement to make sense to the reader. – al-Hwarizmi Aug 24 '13 at 15:20

1 Answers1

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HINT:

Using this, $$(a^n-1,a^m-1)=a^{(n,m)}-1$$

$$\implies (a^m-1)|(a^n-1)\iff m|n$$

$$\implies (x^n-1)|(x^{2012}-1)\iff n|2012$$

Now, the factors of $2012 $ are $n=1,2,4,503,1006,2012$

Observe that $\displaystyle\frac{x^{2012}-1}{x^n-1}$ must have $(x+1)$ as factor

and we know from (1) of this, $ x^m=1$ does not have a repeated root

If $n$ is even, $(x^{2012}-1,x^n-1)$ will be divisible by $x^2-1,$ hence by $x+1$

$\implies (x+1)$ will not divide $\displaystyle\frac{x^{2012}-1}{x^n-1}$

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lab bhattacharjee
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