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 "2023 AMC 10B Problems/Problem 14"

m (i had a bad day that day ok my writing was tripping)
m (→‎Solution 4 (Nice Substitution))
(6 intermediate revisions by 4 users not shown)
Line 17: Line 17:
 
Essentially, this says that the product of two consecutive numbers <math>mn,mn+1</math> must be a perfect square. This is practically impossible except <math>mn=0</math> or <math>mn+1=0</math>.  
 
Essentially, this says that the product of two consecutive numbers <math>mn,mn+1</math> must be a perfect square. This is practically impossible except <math>mn=0</math> or <math>mn+1=0</math>.  
 
<math>mn=0</math> gives <math>(0,0)</math>.
 
<math>mn=0</math> gives <math>(0,0)</math>.
−
<math>mn=-1</math> gives <math>(1,-1), (-1,1)</math>. Answer: <math>\boxed{\textbf{(C)}3}.</math>
+
<math>mn=-1</math> gives <math>(1,-1), (-1,1)</math>. Answer: <math>\boxed{\textbf{(C) 3}}.</math>
  
 
~Technodoggo ~minor edits by lucaswujc
 
~Technodoggo ~minor edits by lucaswujc
Line 41: Line 41:
  
 
Modulo <math>u</math>, we have <math>v^2 \equiv 0 \pmod{u}</math>.
 
Modulo <math>u</math>, we have <math>v^2 \equiv 0 \pmod{u}</math>.
−
Because <math>\left( u, v \right) = 1</math>, we must have <math>|u| = |v| = 1</math>.
+
Because <math>{\rm gcd} \left( u, v \right) = 1</math>., we must have <math>|u| = |v| = 1</math>.
 
Plugging this into the above equation, we get <math>2 + uv = k^2</math>.
 
Plugging this into the above equation, we get <math>2 + uv = k^2</math>.
 
Thus, we must have <math>uv = -1</math> and <math>k = 1</math>.
 
Thus, we must have <math>uv = -1</math> and <math>k = 1</math>.
Line 51: Line 51:
  
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 +
~ sravya_m18
  
 
==Solution 3 (Discriminant)==
 
==Solution 3 (Discriminant)==
−
We can move all terms to one side and wrote the equation as a quadratic in terms of <math>n</math> to get <cmath>(1-m^2)n^2+(m)n+(m^2)=0.</cmath> The discriminant of this quadratic is <cmath>\Delta = m^2-4(1-m^2)(m^2)=m^2(4m^2-3).</cmath> For <math>n</math> to be an integer, we must have <math>m^2(4m^2-3)</math> be a perfect square. Thus, either <math>4m^2-3</math> is a perfect square or <math>m^2 = 0</math> and <math>m = 0</math>. The first case gives <math>m=-1,1</math>, which result in the equations <math>-n+1=0</math> and <math>n-1=0</math>, for a total of two pairs: <math>(-1,1)</math> and <math>(1,-1)</math>. The second case gives the equation <math>n^2=0</math>, so it's only pair is <math>(0,0)</math>. In total, the total number of solutions is <math>\boxed{\textbf{(C) 3}}</math>.
+
We can move all terms to one side and write the equation as a quadratic in terms of <math>n</math> to get <cmath>(1-m^2)n^2+(m)n+(m^2)=0.</cmath> The discriminant of this quadratic is <cmath>\Delta = m^2-4(1-m^2)(m^2)=m^2(4m^2-3).</cmath> For <math>n</math> to be an integer, we must have <math>m^2(4m^2-3)</math> be a perfect square. Thus, either <math>(2m)^2-3</math> is a perfect square or <math>m^2 = 0</math> and <math>m = 0</math>. The first case gives <math>m=-1,1</math> (larger squares are separated by more than 3), which result in the equations <math>-n+1=0</math> and <math>n-1=0</math>, for a total of two pairs: <math>(-1,1)</math> and <math>(1,-1)</math>. The second case gives the equation <math>n^2=0</math>, so it's only pair is <math>(0,0)</math>. In total, the total number of solutions is <math>\boxed{\textbf{(C) 3}}</math>.
  
 
~A_MatheMagician
 
~A_MatheMagician
  
 
==Solution 4 (Nice Substitution)==
 
==Solution 4 (Nice Substitution)==
−
Let <math>x=m+n, y=mn</math> then <cmath>x^2-y=y^2</cmath> Completing the square then gives <cmath>4x^2+1=(2y+1)^2</cmath> Since the RHS is a square, clearly the only solutions are <math>x=0,y=0</math> and <math>x=0,y=-1</math>. The first gives <math>(0,0)</math> while the second gives <math>(-1,1)</math> and <math>(1,-1)</math> by solving it as a quadratic with roots <math>m</math> and <math>n</math>. Thus there are <math>\boxed{\textbf{(C) 3}}</math> solutions.
+
Let <math>x=m+n, y=mn</math> then  
 +
<cmath>x^2-y=y^2</cmath>
 +
 
 +
Completing the square in <math>y</math> and multiplying by 4 then gives
 +
<cmath>4x^2+1=(2y+1)^2</cmath>  
 +
 
 +
Since the RHS is a square, clearly the only solutions are <math>x=0,y=0</math> and <math>x=0,y=-1</math>.
 +
 
 +
The first gives <math>(0,0)</math>.
 +
 
 +
The second gives <math>(-1,1)</math> and <math>(1,-1)</math> by solving it as a quadratic with roots <math>m</math> and <math>n</math>.  
 +
 
 +
Thus there are <math>\boxed{\textbf{(C) 3}}</math> solutions.
  
 
~ Grolarbear
 
~ Grolarbear
Line 70: Line 83:
  
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 +
==Video Solution by Interstigation==
 +
 +
https://youtu.be/Vq7kevsWlHk
  
 +
~Interstigation
  
 
==See also==
 
==See also==

Revision as of 21:31, 9 July 2024

Problem

How many ordered pairs of integers $(m, n)$ satisfy the equation $m^2+mn+n^2 = m^2n^2$?

$\textbf{(A) }7\qquad\textbf{(B) }1\qquad\textbf{(C) }3\qquad\textbf{(D) }6\qquad\textbf{(E) }5$

Solution 1

Clearly, $m=0,n=0$ is one of the solutions. However, we can be quite sure that there are more, so we apply Simon's Favorite Factoring Trick to get the following:

\begin{align*} m^2+mn+n^2 &= m^2n^2\\ m^2+mn+n^2 +mn &= m^2n^2 +mn\\ (m+n)^2 &= m^2n^2 +mn\\ (m+n)^2 &= mn(mn+1).\\ \end{align*}

Essentially, this says that the product of two consecutive numbers $mn,mn+1$ must be a perfect square. This is practically impossible except $mn=0$ or $mn+1=0$. $mn=0$ gives $(0,0)$. $mn=-1$ gives $(1,-1), (-1,1)$. Answer: $\boxed{\textbf{(C) 3}}.$

~Technodoggo ~minor edits by lucaswujc

Solution 2

Case 1: $mn = 0$.

In this case, $m = n = 0$.

Case 2: $mn \neq 0$.

Denote $k = {\rm gcd} \left( m, n \right)$. Denote $m = k u$ and $n = k v$. Thus, ${\rm gcd} \left( u, v \right) = 1$.

Thus, the equation given in this problem can be written as \[ u^2 + uv + v^2 = k^2 u^2 v^2 . \]

Modulo $u$, we have $v^2 \equiv 0 \pmod{u}$. Because ${\rm gcd} \left( u, v \right) = 1$., we must have $|u| = |v| = 1$. Plugging this into the above equation, we get $2 + uv = k^2$. Thus, we must have $uv = -1$ and $k = 1$.

Thus, there are two solutions in this case: $\left( m , n \right) = \left( 1, -1 \right)$ and $\left( m , n \right) = \left( -1, 1 \right)$.

Putting all cases together, the total number of solutions is $\boxed{\textbf{(C) 3}}$.

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) ~ sravya_m18

Solution 3 (Discriminant)

We can move all terms to one side and write the equation as a quadratic in terms of $n$ to get \[(1-m^2)n^2+(m)n+(m^2)=0.\] The discriminant of this quadratic is \[\Delta = m^2-4(1-m^2)(m^2)=m^2(4m^2-3).\] For $n$ to be an integer, we must have $m^2(4m^2-3)$ be a perfect square. Thus, either $(2m)^2-3$ is a perfect square or $m^2 = 0$ and $m = 0$. The first case gives $m=-1,1$ (larger squares are separated by more than 3), which result in the equations $-n+1=0$ and $n-1=0$, for a total of two pairs: $(-1,1)$ and $(1,-1)$. The second case gives the equation $n^2=0$, so it's only pair is $(0,0)$. In total, the total number of solutions is $\boxed{\textbf{(C) 3}}$.

~A_MatheMagician

Solution 4 (Nice Substitution)

Let $x=m+n, y=mn$ then \[x^2-y=y^2\]

Completing the square in $y$ and multiplying by 4 then gives \[4x^2+1=(2y+1)^2\]

Since the RHS is a square, clearly the only solutions are $x=0,y=0$ and $x=0,y=-1$.

The first gives $(0,0)$.

The second gives $(-1,1)$ and $(1,-1)$ by solving it as a quadratic with roots $m$ and $n$.

Thus there are $\boxed{\textbf{(C) 3}}$ solutions.

~ Grolarbear

Video Solution by OmegaLearn

https://youtu.be/5a5caco_YTo

Video Solution

https://youtu.be/Dh1lDI1fHrw

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Video Solution by Interstigation

https://youtu.be/Vq7kevsWlHk

~Interstigation

See also

2023 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
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
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AMC_10B_Problems/Problem_14&oldid=221990"