1980 AHSME Problems/Problem 17

Problem

Given that $i^2=-1$ , for how many integers $n$ is $(n+i)^4$ an integer?

$\text{(A)} \ \text{none} \qquad \text{(B)} \ 1 \qquad \text{(C)} \ 2 \qquad \text{(D)} \ 3 \qquad \text{(E)} \ 4$

Solution 1

Expanding $(n+i)^4$ yields $n^4+4in^3-6n^2-4in+1 = (n^4-6n^2+1) + (4n^3-4n)i$ . This quantity is an integer if and only if $4n^3 - 4n = 4n(n+1)(n-1) = 0$ , that is, if $n \in \{0, 1, -1\}$ . Therefore, there are $\boxed{(\textbf{D})\ 3}$ such values of $n$ .

-aopspandy, edited by j314andrews

Solution 2

For $(n+i)^4$ to be a real number, $n+i$ must be a scalar multiple of an eighth root of unity, that is, $n+i = r\left(\cos \frac{k\pi}{4} + i \sin \frac{k\pi}{4}\right)$ where $r \geq 0$ is a real number and $k$ is an integer such that $0 \leq k \leq 7$ . So $r \sin \frac{k\pi}{4} = 1$ and $\sin \frac{k\pi}{4}$ is positive. Therefore, $k \in \{1, 2, 3\}$ .

If $k = 1$ , $r = \frac{1}{\sin\frac{\pi}{4}} = \sqrt{2}$ , $n = \sqrt{2}\cos\frac{\pi}{4} = 1$ and $(n+i)^4 = (1+i)^4 = -4$ .

If $k = 2$ , $r = \frac{1}{\sin\frac{\pi}{2}} = 1$ , $n = \cos\frac{\pi}{2} = 0$ and $(n+i)^4 = i^4 = 1$ .

If $k = 3$ , $r = \frac{1}{\sin\frac{3\pi}{4}} = \sqrt{2}$ , $n = \sqrt{2}\cos\frac{3\pi}{4} = -1$ and $(n+i)^4 = (-1+i)^4 = -4$ .

In all these cases, $(n+i)^4$ is an integer, so $\{1, 0, -1\}$ are the $\boxed{(\mathbf{D})\ 3}$ possible values of $n$ .

~ jaspersun, edited by j314andrews

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1980 AHSME Problems/Problem 17

Contents

Problem

Solution 1

Solution 2

See also