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

$(n+i)^4=n^4+4in^3-6n^2-4in+1$ , and this has to be an integer, so the sum of the imaginary parts must be $0$ . $4in^3-4in=0$ $4in^3=4in$ $n^3=n$ Since $n^3=n$ , there are $\boxed{3}$ solutions for $n$ : $0$ and $\pm1$ .

-aopspandy

Solution

Since we have an imaginary term, we can think about rotations. We are in the first and second quadrant, so we only need to think about angles from 0 to $\pi$ exclusive. Specifically, $4\theta = \pi k$ , where $k$ is an integer. Therefore, the only angles which can work are $\pi/4, \pi/2$ and $3\pi/4$ .

Now we just need to see if these angles can be represented by $(n+i)^4$ . $\pi/4$ and $3\pi/4$ work, since they form a 45-45-90 triangle, and $\pi/2$ works, since it doesn't have a real component.

So, the answer is $\boxed{D}$ .

~ jaspersun

1980 AHSME (Problems • Answer Key • Resources)
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1980 AHSME Problems/Problem 17

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