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Difference between revisions of "1980 AHSME Problems/Problem 17"
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== Solution ==
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== Solution 1 ==
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<math>(n+i)^4=n^4+4in^3-6n^2-4in+1</math>, and this has to be an integer, so the sum of the imaginary parts must be <math>0</math>. <cmath>4in^3-4in=0</cmath>  <cmath> 4in^3=4in</cmath> <cmath>n^3=n</cmath>
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Expanding <math>(n+i)^4</math> yields <math>n^4+4in^3-6n^2-4in+1 = (n^4-6n^2+1) + (4n^3-4n)i</math>.  This quantity is an integer if and only if <math>4n^3 - 4n = 4n(n+1)(n-1) = 0</math>, that is, if <math>n \in \{0, 1, -1\}</math>.  Therefore, there are <math>\boxed{(\textbf{D})\ 3}</math> such values of <math>n</math>.
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Since <math>n^3=n</math>, there are <math>\boxed{3}</math> solutions for <math>n</math>: <math>0</math> and <math>\pm1</math>.
 
  
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-aopspandy
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-aopspandy, edited by j314andrews
  
 
== Solution ==
 
== Solution ==

Revision as of 12:39, 15 August 2025

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

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 (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
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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