There's only 2023066 cases to check

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.

#Algebra #Sharky

Easy Math Editor

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Comments

Very 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 - 5 years, 7 months ago

I'm getting $671^{2}$ polynomials.

Julian Poon - 5 years, 7 months ago

Can you give proof?

Sharky Kesa - 5 years, 7 months ago

Ok.

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 - 5 years, 7 months ago

@Julian Poon – Nice solution. Now list them all!!! (Mwa-ha-ha-ha-ha) :P

Sharky Kesa - 5 years, 7 months ago

@Julian Poon – Well I did not understand a single thing please elaborate (from $r_{1}^{n}$ line, please!)

Department 8 - 5 years, 7 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$