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

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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>
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

Latest revision as of 19:10, 18 June 2021

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

See also

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