# 1980 AHSME Problems/Problem 28

## Problem

The polynomial $x^{2n}+1+(x+1)^{2n}$ is not divisible by $x^2+x+1$ if $n$ equals $\text{(A)} \ 17 \qquad \text{(B)} \ 20 \qquad \text{(C)} \ 21 \qquad \text{(D)} \ 64 \qquad \text{(E)} \ 65$

## Solution 1

Let $h(x)=x^2+x+1$.

Then we have $$(x+1)^2n = (x^2+2x+1)^n = (h(x)+x)^n = g(x) \cdot h(x) + x^n,$$ where $g(x)$ is $h^{n-1}(x) + nh^{n-2}(x) \cdot x + ... + x^{n-1}$ (after expanding $(h(x)+x)^n$ according to the Binomial Theorem).

Notice that $$x^2n = x^2n+x^{2n-1}+x^{2n-2}+...x -x^{2n-1}-x^{2n-2}-x^{2n-3} +...$$ $x^n = x^n+x^{n-1}+x^{n-2} -x^{n-1}-x^{n-2}-x^{n-3} +....$

Therefore, the left term from $x^2n$ is $x^{(2n-3u)}$

          the left term from $x^n$ is $x^{(n-3v)}$,


If divisible by h(x), we need 2n-3u=1 and n-3v=2 or

                             2n-3u=2 and n-3v=1


The solution will be n=1 or 2 mod(3). Therefore n=21 is impossible

~~Wei

## Solution 2

Notice that the roots of $w^2+w+1=0$ are also the third roots of unity (excluding $w=1$). This is fairly easy to prove: multiply both sides by $w-1$ and we get $$(w-1)(w^2+w+1) = w^3 - 1 = 0.$$ These roots are $w = e^{i \pi /3}$ and $w = e^{2i \pi /3}$.

Now we have \begin{align*} w^{2n} + 1 + (w+1)^{2n} &= w^{2n} + 1 + (-w^2)^{2n} \\ &= w^{4n} + w^{2n} + 1\\ &= 0. \end{align*} Plug in the roots of $w^2+w+1=0$. Note that $$e^{2i \pi /3} + 1 = -\frac{1}{2} + \frac{\sqrt{3}}{2}i + 1 = \frac{1}{2} + \frac{\sqrt{3}}{2}i = e^{i \pi /3}.$$ However, this will not work if $n=3m$, so $n$ cannot be equal to $21$. Hence our answer is $\textrm{(C)}$.

## Solution 3

We start by noting that $$x + 1 \equiv -x^2 \mod (x^2+x+1).$$ Let $n = 3k+r$, where $r \in \{ 0,1,2 \}$.

Thus we have $$x^{4n} + x^{2n} + 1 \equiv x^{4r} + x^{2r} + 1 \mod (x^3 -1).$$

When $r = 0$, $$x^{4n} + x^{2n} + 1 \equiv 3 \mod (x^3 -1).$$ When $r = 1$, $$x^{4n} + x^{2n} + 1 \equiv x^2 + x + 1 \mod (x^3 -1),$$ which will be divisible by $x^2+x+1$.

When $r = 2$, $$x^{4n} + x^{2n} + 1 \equiv x^2 + x + 1 \mod (x^3 -1),$$ which will also be divisible by $x^2+x+1$.

Thus $r \ne 0$, so $n$ cannot be divisible by $3$, and the answer is $\textrm{(C)}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 