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Difference between revisions of "1980 AHSME Problems/Problem 17"

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<math>\text{(A)} \ \text{none} \qquad \text{(B)} \ 1 \qquad \text{(C)} \ 2 \qquad \text{(D)} \ 3 \qquad \text{(E)} \ 4</math>
 
<math>\text{(A)} \ \text{none} \qquad \text{(B)} \ 1 \qquad \text{(C)} \ 2 \qquad \text{(D)} \ 3 \qquad \text{(E)} \ 4</math>
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== Solution ==
 
== Solution ==
 
<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>
 
<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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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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Since <math>n^3=n</math>, there are <math>\boxed{3}</math> solutions for <math>n</math>: <math>0</math> and <math>\pm1</math>.
  
 
-aopspandy
 
-aopspandy
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 +
== Solution ==
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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 <math>\pi</math> exclusive. Specifically, <math>4\theta = \pi k</math>, where <math>k</math> is an integer. Therefore, the only angles which can work are <math>\pi/4, \pi/2</math> and <math>3\pi/4</math>.
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Now we just need to see if these angles can be represented by <math>(n+i)^4</math>. <math>\pi/4</math> and <math>3\pi/4</math> work, since they form a 45-45-90 triangle, and <math>\pi/2</math> works, since it doesn't have a real component.
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So, the answer is <math>\boxed{D}</math>.
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~ jaspersun
  
 
== See also ==
 
== See also ==

Latest revision as of 14:38, 10 September 2024

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 (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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