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The number of natural number $n$ in the interval of [$1005,2010$ ] for which the polynomial $1$+$x$+$x^2$+ $x^3$ +...+$x^{n-1}$ divides the polynomial $1$+$x^2$+$x^4$+$x^6$ ....+$x^{2010}$ is

Ans is the number of such numbers is 503.

My soln: $$ (x^{503}-1)(x^{503} +1)( x^{1006}+1) $$ to be Divided by $ x^n -1$ So n=503.

maveric
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    Contest problem from which contest in 2010? Please cite the source of such problems statements. – amWhy May 07 '20 at 21:04
  • What have you tried? – Wojowu May 07 '20 at 21:05
  • It is kvpy 2010 exam. I am gettin n=503, but the question and its official answer key state answer as 503 as number of numbers within the interval as 503. I need to check so what is suppose to be correct answer. – maveric May 07 '20 at 21:06
  • @JohnOmielan The answer should be the number of such numbers, so 503 makes sense. However, I have a strong suspicion this is not the correct answer. – Wojowu May 07 '20 at 21:06
  • @Wojowu Thanks, I misread the question. – John Omielan May 07 '20 at 21:07
  • I did write it as $x^{503}-1 {1-x+x^2.....}{ 1+x^{1006}$ which now needs to be divided by $x^{n}-1$ – maveric May 07 '20 at 21:09
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    If we have the asked divisibility of polynomials, then, since they are monic, the quotient must be an integer polynomial. In particular, we must also have a divisibility of values: for $x=1$, we get that $n$ should divide $1006$, and in the given interval there is only one such value. – Wojowu May 07 '20 at 21:10
  • What value is that? – maveric May 07 '20 at 21:12
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    https://math.stackexchange.com/questions/475084/natural-number-n-divisibility – lab bhattacharjee May 08 '20 at 02:09
  • @bhattachargee i asked same q there also? So the given official answer key was wrong? – maveric May 08 '20 at 07:03

2 Answers2

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The polynomial $1+x+x^2+...+x^{n-1}$ divides $1+x^2+x^4+...+x^{2010}$ exactly when $$ (x^n-1)=(x-1)(1+x+...+x^{n-1})|(x-1)(1+x^2+x^4+...)=\frac{x^2-1}{x+1}(1+x^2+...+x^{2010})=\frac{x^{2012}-1}{x+1}. $$

If the polynomial $x^n-1$ divides the polynomial $\frac{x^{2012}-1}{x+1},$ then all the (complex) roots of $x^n-1$ must also be roots of $x^{2012}-1,$ but $x=-1$ cannot be a root of $x^n-1.$. Because $-1$ is a root of $x^n-1$ when $n$ is even, the value $n$ must be odd. If $n$ is odd, then it is sufficient that $x^n-1|x^{2012}-1.$ But the complex roots of $x^n-1$ are shared by $x^{2012}-1$ exactly when $n$ divides $2012$. The only value $n$ in the range $[1005, 2010]$ for which this happens is $1006$, but this even, so there are zero solutions for $n$ in the range $[1005,2010].$

subrosar
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  • Why dont your solution considers n=503? For n=503 second polynomial does divide first . Also common ratio of one of the terms of x^2 – maveric May 07 '20 at 21:34
  • Sorry, I misread the question the first time I posted, and then I had some errors to fix. Now my solution shows that the first polynomial divides the second when $n$ is any odd divisor of $2012.$ But since $1006$ is even and not odd, it doesn't make the cut. – subrosar May 07 '20 at 22:00
  • How odd n elliminated? Also n=503 is valid solution. – maveric May 07 '20 at 22:35
  • The even $n$ are eliminated. The odd $n$ are valid solutions. If $n$ is even then $-1$ is a root of $1+x+...+x^{n-1}$ so it contains $(x+1)$ as a factor. But $-1$ is not a root of $1+x^2+x^4+...+x^{2010}$ so this polynomial does not contain $(1+x)$ as a factor. Therefore if $n$ is even the first polynomial cannot divide the second. – subrosar May 07 '20 at 23:32
  • @maveric $n = 503$ would be a solution except it is not in $[1005, 2010]$ – Doug M May 07 '20 at 23:59
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If $f(x)=1+x+x^2+\cdots+x^{n-1}$ divides $g(x)=1+x^2+x^4+\cdots+x^{2010}$, then the quotient has integer coefficients. Therefore, $n=f(1)$ divides $1006=g(1)$. There is only one multiple of $1006$ in $[1005,2010]$, which is $1006$ itself. Thus, the only possible solution, if there is a solution, is $n=1006$.

lhf
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    I'm not sure why people downvoted this response. Seems perfectly legit to me, even if it does not resolve the question of whether or not $n=1006$ is a solution (it is not). – subrosar May 08 '20 at 16:30