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1980 AHSME Problems/Problem 28

Revision as of 19:53, 17 August 2025 by J314andrews (talk | contribs) (→‎Solution 2)

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

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 2

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 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 $\textrm{(C)}$.

See also

1980 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 27 		Followed by
Problem 29
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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