How many polynomials are there of the form \(x^m + x^n + 1\), where \(m\) and \(n\) are positive integers such that \(1 \leq m < n \leq 2012\), that are divisible by \(x^2 + x + 1\)? Give proof.

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TopNewestVery simply, substitute x=ω. This gives n=3k+1 and m=3k+2 and vice-versa. Now simply count the number of cases. – Yugesh Kothari · 1 year, 2 months ago

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I'm getting \(671^{2}\) polynomials. – Julian Poon · 1 year, 2 months ago

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– Sharky Kesa · 1 year, 2 months ago

Can you give proof?Log in to reply

If \(f(x)=x^{m} + x^{n} + 1\)is divisible by \(g(x)=x^{2} + x+1\), then the roots of \(g(x)\) must also be the roots of \(f(x)\). Or that \(f(r_{1})=f(r_{2})=0\) where \(r_{1}\) and \(r_{2}\) are the roots of \(g(x)\).

Since in order for \(f(r_{1})=f(r_{2})=0\), we must find some value of \(n\) and \(m\) to be such that \(r_1^n+r_1^m=-1\) and \(r_2^n+r_2^m=-1\)

The roots of \(g(x)\) are \[r_1=-\left(-1\right)^{\frac{1}{3}}=-\frac{1}{2}-\frac{i\sqrt{3}}{2}\] \[r_2=\left(-1\right)^{\frac{2}{3}}=-\frac{1}{2}+\frac{i\sqrt{3}}{2}\]

It can be seen that \(r_1^n=r_1^{n+3k}\) and \(r_2^n=r_2^{n+3k}\) where \(n\) and \(k\) are some integers. And that \(r_1+r_1^2=r_2+r_2^2=-1\)

Hence, either \(m\) or \(n\) has to be of the form \(3k+1\) while the other has to be of \(3k+2\), where \(k\) is an integer.

The number of possible values of \(m\) and \(n\) that satisfy the boundaries \(1\leq m < n \leq 2012\) can be calculated to be \(671^{2}\) – Julian Poon · 1 year, 2 months ago

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– Sharky Kesa · 1 year, 2 months ago

Nice solution. Now list them all!!! (Mwa-ha-ha-ha-ha) :PLog in to reply

– Lakshya Sinha · 1 year, 2 months ago

Well I did not understand a single thing please elaborate (from \(r_{1}^{n}\) line, please!)Log in to reply