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

1980 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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1980 AHSME Problems/Problem 17

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