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

See also

1980 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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