Difference between revisions of "1980 AHSME Problems/Problem 28"

Latest revision as of 20:03, 17 August 2025

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

The roots of $x^2+x+1$ are $-\frac{1}{2} + \frac{\sqrt{3}}{2}i = \cos\frac{2}{3}\pi + i \sin \frac{2}{3}\pi$ and $-\frac{1}{2} - \frac{\sqrt{3}}{2}i = \cos \frac{4}{3}\pi + i \sin\frac{4}{3}\pi$ , which are the primitive third roots of unity.

Let $w = -\frac{1}{2} + \frac{\sqrt{3}}{2}i = \cos\frac{2}{3}\pi + i \sin \frac{2}{3}\pi$ . Note that $-\frac{1}{2} - \frac{\sqrt{3}}{2}i = \cos \frac{4}{3}\pi + i \sin\frac{4}{3}\pi = w^2 = \bar{w}$ . So by the conjugate root theorem, $x^{2n} + 1 + (1+x)^{2n}$ is divisible by $x^2+x+1$ if and only if $w$ is a root of $x^{2n} + 1 + (1+x)^{2n}$ .

Also, $w^{2n} + 1 + (w+1)^{2n} = w^{2n} + 1 + (-w^2)^{2n} = w^{2n} + 1 + w^{4n} = w^{2n} + 1 + w^n = w^{2n} + w^n + 1$ .

If $n \equiv 0\ (\textrm{mod}\ 3)$ , $w^{2n} + w^n + 1 \equiv 1 + 1 + 1 = 3$ .

If $n \equiv 1\ (\textrm{mod}\ 3)$ , $w^{2n} + w^n + 1 \equiv w^2 + w + 1 = 0$ .

If $n \equiv 2\ (\textrm{mod}\ 3)$ , $w^{2n} + w^n + 1 \equiv w + w^2 + 1 = 0$ .

Therefore, $x^{2n} + 1 + (1+x)^{2n}$ is not divisible by $x^2 + x + 1$ if and only if $n$ is divisible by $3$ . Of the answer choices, only $\boxed{(\textbf{C})\ 21}$ is divisible by $3$ .

Solution 2

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

Then $(x+1)^{2n} = (x^2+2x+1)^n = (x+h(x))^n = \sum_{k=0}^{n} \binom{n}{k} x^k (h(x))^{n-k}$ $= \sum_{k=0}^{n-1} \binom{n}{k} x^k (h(x))^{n-k} + x^n = h(x) \left(\sum_{k=0}^{n-1} \binom{n}{k} x^k (h(x))^{n-k-1}\right) + x^n$ .

Let $g(x) = \sum_{k=0}^{n-1} \binom{n}{k} x^k (h(x))^{n-k-1}$ . Then $x^{2n} + 1 + (x+1)^{2n} = x^{2n} + 1 + g(x)h(x) + x^n$ , so $x^{2n} + 1 + (x+1)^{2n}$ is divisible by $x^2 + x + 1$ if and only if $x^{2n} + x^n + 1$ is divisible by $x^2 + x + 1$ .

Let $k$ be any non-negative integer, and let $r \in \{0, 1, 2\}$ . Then $x^{3k+r} - x^r = x^r(x^{3k} - 1) = x^r(x^3 - 1)\left(\sum_{j=0}^{k - 1} x^{3j}\right)$ $= x^r(x-1)(x^2 + x + 1)\left(\sum_{j=0}^{k - 1} x^{3j}\right)$ .

Therefore, $x^{3k+r} - x^r$ is divisible by $x^2 + x + 1$ .

If $n$ is a nonnegative integer, then $n$ can be written as $3k + r$ , where nonnegative integer $k$ and $r \in \{0, 1, 2\}$ are the quotient and remainder when $n$ is divided by $3$ .

If $r = 2$ , then $x^{2n} + x^n + 1 = x^{6k+4} + x^{3k+2} + 1 = x^{3(2k+1)+1} + x^{3k+2} + 1 = (x^{3(2k+1)+1} - x) + (x^{3k+2} - x^2) + x^2 + x + 1$ . Since $x^{3(2k+1)+1} - x$ and $(x^{3k+2} - x^2)$ are divisible by $x^2 + x + 1$ , $x^{2n} + x^n + 1$ is divisible by $x^2 + x + 1$ .

If $r = 1$ , then $x^{2n} + x^n + 1 = x^{6k+2} + x^{3k+1} + 1 = x^{3(2k)+2} + x^{3k+1} + 1 = (x^{3(2k)+2} - x^2) + (x^{3k+1} - x) + x^2 + x + 1$ . Since $x^{3(2k)+2} - x^2$ and $(x^{3k+1} - x)$ are divisible by $x^2 + x + 1$ , $x^{2n} + x^n + 1$ is divisible by $x^2 + x + 1$ .

If $r = 0$ , then $x^{2n} + x^n + 1 = x^{6k} + x^{3k} + 1 = (x^{6k} - 1) + (x^{3k} - 1) + 3$ . Since $x^{6k} - 1$ and $x^{3k} - 1$ are divisible by $x^2 + x + 1$ , $x^{2n} + x^n + 1$ is not divisible by $x^2 + x + 1$ .

So $x^{2n} + 1 + (1+x)^{2n}$ is not divisible by $x^2 + x + 1$ if and only if $r = 0$ , that is, $n$ is divisible by $3$ . The only answer choice that is divisible by $3$ is $\boxed{(\textbf{C})\ 21}$ .

-Wei, edited by j314andrews

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 $\textrm{(C)}$ .

@@ Line 9: / Line 9: @@
 \text{(E)} \ 65  </math>
-== Solution 1==
+==Solution 1==
+The roots of <math>x^2+x+1</math> are <math>-\frac{1}{2} + \frac{\sqrt{3}}{2}i = \cos\frac{2}{3}\pi + i \sin \frac{2}{3}\pi</math> and <math>-\frac{1}{2} - \frac{\sqrt{3}}{2}i = \cos \frac{4}{3}\pi + i \sin\frac{4}{3}\pi</math>, which are the primitive third roots of unity.
+Let <math>w = -\frac{1}{2} + \frac{\sqrt{3}}{2}i = \cos\frac{2}{3}\pi + i \sin \frac{2}{3}\pi</math>.  Note that <math>-\frac{1}{2} - \frac{\sqrt{3}}{2}i = \cos \frac{4}{3}\pi + i \sin\frac{4}{3}\pi = w^2 = \bar{w}</math>.  So by the conjugate root theorem, <math>x^{2n} + 1 + (1+x)^{2n}</math> is divisible by <math>x^2+x+1</math> if and only if <math>w</math> is a root of <math>x^{2n} + 1 + (1+x)^{2n}</math>.
+Also, <math>w^{2n} + 1 + (w+1)^{2n} = w^{2n} + 1 + (-w^2)^{2n} = w^{2n} + 1 + w^{4n} = w^{2n} + 1 + w^n = w^{2n} + w^n + 1</math>.
+If <math>n \equiv 0\ (\textrm{mod}\ 3)</math>, <math>w^{2n} + w^n + 1 \equiv 1 + 1 + 1 = 3</math>.
+If <math>n \equiv 1\ (\textrm{mod}\ 3)</math>, <math>w^{2n} + w^n + 1 \equiv w^2 + w + 1 = 0</math>.
+If <math>n \equiv 2\ (\textrm{mod}\ 3)</math>, <math>w^{2n} + w^n + 1 \equiv w + w^2 + 1 = 0</math>.
+Therefore, <math>x^{2n} + 1 + (1+x)^{2n}</math> is not divisible by <math>x^2 + x + 1</math> if and only if <math>n</math> is divisible by <math>3</math>.  Of the answer choices, only <math>\boxed{(\textbf{C})\  21}</math> is divisible by <math>3</math>.
+== Solution 2==
 Let <math>h(x)=x^2+x+1</math>.
@@ Line 32: / Line 47: @@
 -Wei, edited by j314andrews
-==Solution 2==
-The roots of <math>x^2+x+1</math> are <math>-\frac{1}{2} + \frac{\sqrt{3}}{2}i = \cos\frac{2}{3}\pi + i \sin \frac{2}{3}\pi</math> and <math>-\frac{1}{2} - \frac{\sqrt{3}}{2}i = \cos \frac{4}{3}\pi + i \sin\frac{4}{3}\pi</math>, which are the primitive third roots of unity.
-Let <math>w = -\frac{1}{2} + \frac{\sqrt{3}}{2}i = \cos\frac{2}{3}\pi + i \sin \frac{2}{3}\pi</math>.  Note that <math>-\frac{1}{2} - \frac{\sqrt{3}}{2}i = \cos \frac{4}{3}\pi + i \sin\frac{4}{3}\pi = w^2 = \bar{w}</math>.  So by the conjugate root theorem, <math>x^{2n} + 1 + (1+x)^{2n}</math> is divisible by <math>x^2+x+1</math> if and only if <math>w</math> is a root of <math>x^{2n} + 1 + (1+x)^{2n}</math>.
-Also, <math>w^{2n} + 1 + (w+1)^{2n} = w^{2n} + 1 + (-w^2)^{2n} = w^{2n} + 1 + w^{4n} = w^{2n} + 1 + w^n = w^{2n} + w^n + 1</math>.
-If <math>n \equiv 0\ (\textrm{mod}\ 3)</math>, <math>w^{2n} + w^n + 1 \equiv 1 + 1 + 1 = 3</math>.
-If <math>n \equiv 1\ (\textrm{mod}\ 3)</math>, <math>w^{2n} + w^n + 1 \equiv w^2 + w + 1 = 0</math>.
-If <math>n \equiv 2\ (\textrm{mod}\ 3)</math>, <math>w^{2n} + w^n + 1 \equiv w + w^2 + 1 = 0</math>.
-Therefore, <math>x^{2n} + 1 + (1+x)^{2n}</math> is not divisible by <math>x^2 + x + 1</math> if and only if <math>n</math> is divisible by <math>3</math>.  Of the answer choices, only <math>\boxed{(\textbf{C})\  21}</math> is divisible by <math>3</math>.
 ==Solution 3==

1980 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 27	Followed by Problem 29
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