Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution)
(→‎Solution 2)
 
(5 intermediate revisions by 2 users not shown)
Line 6: Line 6:
  
  
−
== 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>
 
−
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
+
== Solution 1 ==
 +
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>.
 +
 
 +
-aopspandy, edited by j314andrews
 +
 
 +
== Solution 2 ==
 +
For <math>(n+i)^4</math> to be a real number, <math>n+i</math> must be a scalar multiple of an eighth root of unity, that is, <math>n+i = r\left(\cos \frac{k\pi}{4} + i \sin \frac{k\pi}{4}\right)</math> where <math>r \geq 0</math> is a real number and <math>k</math> is an integer such that <math>0 \leq k \leq 7</math>.  So <math>r \sin \frac{k\pi}{4} = 1</math> and <math>\sin \frac{k\pi}{4}</math> is positive.  Therefore, <math>k \in \{1, 2, 3\}</math>.
 +
 
 +
If <math>k = 1</math>, <math>r = \frac{1}{\sin\frac{\pi}{4}} = \sqrt{2}</math>, <math>n = \sqrt{2}\cos\frac{\pi}{4} = 1</math> and <math>(n+i)^4 = (1+i)^4 = -4</math>.
 +
 
 +
If <math>k = 2</math>, <math>r = \frac{1}{\sin\frac{\pi}{2}} = 1</math>, <math>n = \cos\frac{\pi}{2} = 0</math> and <math>(n+i)^4 = i^4 = 1</math>. 
 +
 
 +
If <math>k = 3</math>, <math>r = \frac{1}{\sin\frac{3\pi}{4}} = \sqrt{2}</math>, <math>n = \sqrt{2}\cos\frac{3\pi}{4} = -1</math> and <math>(n+i)^4 = (-1+i)^4 = -4</math>. 
 +
 
 +
In all these cases, <math>(n+i)^4</math> is an integer, so <math>\{1, 0, -1\}</math> are the <math>\boxed{(\mathbf{D})\ 3}</math> possible values of <math>n</math>.
 +
 
 +
~ jaspersun, edited by j314andrews
  
 
== See also ==
 
== See also ==

Latest revision as of 13:07, 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 2

For $(n+i)^4$ to be a real number, $n+i$ must be a scalar multiple of an eighth root of unity, that is, $n+i = r\left(\cos \frac{k\pi}{4} + i \sin \frac{k\pi}{4}\right)$ where $r \geq 0$ is a real number and $k$ is an integer such that $0 \leq k \leq 7$. So $r \sin \frac{k\pi}{4} = 1$ and $\sin \frac{k\pi}{4}$ is positive. Therefore, $k \in \{1, 2, 3\}$.

If $k = 1$, $r = \frac{1}{\sin\frac{\pi}{4}} = \sqrt{2}$, $n = \sqrt{2}\cos\frac{\pi}{4} = 1$ and $(n+i)^4 = (1+i)^4 = -4$.

If $k = 2$, $r = \frac{1}{\sin\frac{\pi}{2}} = 1$, $n = \cos\frac{\pi}{2} = 0$ and $(n+i)^4 = i^4 = 1$.

If $k = 3$, $r = \frac{1}{\sin\frac{3\pi}{4}} = \sqrt{2}$, $n = \sqrt{2}\cos\frac{3\pi}{4} = -1$ and $(n+i)^4 = (-1+i)^4 = -4$.

In all these cases, $(n+i)^4$ is an integer, so $\{1, 0, -1\}$ are the $\boxed{(\mathbf{D})\ 3}$ possible values of $n$.

~ jaspersun, edited by j314andrews

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1980_AHSME_Problems/Problem_17&oldid=257655"