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Difference between revisions of "2023 AMC 10B Problems/Problem 14"

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==Problem==
 
How many ordered pairs of integers <math>(m, n)</math> satisfy the equation <math>m^2+mn+n^2 = m^2n^2</math>?
 
How many ordered pairs of integers <math>(m, n)</math> satisfy the equation <math>m^2+mn+n^2 = m^2n^2</math>?
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<math>\textbf{(A) }7\qquad\textbf{(B) }1\qquad\textbf{(C) }3\qquad\textbf{(D) }6\qquad\textbf{(E) }5</math>
  
 
== Solution 1 ==
 
== Solution 1 ==
  
−
Clearly, <math>m=0,n=0</math> is 1 solution. However there are definitely more, so we apply [https://artofproblemsolving.com/wiki/index.php/Simon%27s_Favorite_Factoring_Trick Simon's Favorite Factoring Expression] to get this:
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Clearly, <math>m=0,n=0</math> is one of the solutions. However, we can be quite sure that there are more, so we apply [https://artofproblemsolving.com/wiki/index.php/Simon%27s_Favorite_Factoring_Trick Simon's Favorite Factoring Trick] to get the following:  
  
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
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m^2+mn+n^2 +mn &= m^2n^2 +mn\\
 
m^2+mn+n^2 +mn &= m^2n^2 +mn\\
 
(m+n)^2 &= m^2n^2 +mn\\
 
(m+n)^2 &= m^2n^2 +mn\\
−
(m+n)^2 &= mn(mn+1)\\
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(m+n)^2 &= mn(mn+1).\\
 
\end{align*}</cmath>
 
\end{align*}</cmath>
  
−
This basically say that the product of two consecutive numbers <math>mn,mn+1</math> must be a perfect square which is practically impossible except <math>mn=0</math> or <math>mn+1=0</math>.
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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>.
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<math>mn=-1</math> gives <math>(1,-1), (-1,1)</math>.
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<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
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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>.
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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>.
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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~ sravya_m18
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==Solution 3 (Discriminant)==
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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>.
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~A_MatheMagician
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==Solution 4 (Nice Substitution)==
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Let <math>x=m+n, y=mn</math> then
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<cmath>x^2-y=y^2</cmath>
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Completing the square in <math>y</math> and multiplying by 4 then gives
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<cmath>4x^2+1=(2y+1)^2</cmath>
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Since the RHS is a square, clearly the only solutions are <math>x=0,y=0</math> and <math>x=0,y=-1</math>.
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The first gives <math>(0,0)</math>.
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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>.
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Thus there are <math>\boxed{\textbf{(C) 3}}</math> solutions.
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~ Grolarbear
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==Solution 5 (Alternative Method for Manipulation)==
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<math>m^2 + mn + n^2 = m^2n^2</math>
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<math>mn = m^2n^2 - m^2 - n^2</math>
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<math>mn + 1 = m^2(n^2 - 1) - 1(n^2 - 1)</math>
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<math>mn + 1 = (m + 1)(m - 1)(n + 1)(n - 1)</math>
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Notice that the right side can be zero or one.
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If the right side is zero, m and n can be <math>(-1,1)</math> and <math>(1,-1)</math>.
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If the right side is one, m and n can be <math>(0,0)</math>.
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There are <math>\boxed{\textbf{(C) 3}}</math> solutions.
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~unhappyfarmer
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==Video Solution by OmegaLearn==
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https://youtu.be/5a5caco_YTo
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==Video Solution==
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 +
https://youtu.be/Dh1lDI1fHrw
 +
 +
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 +
==Video Solution by Interstigation==
 +
 +
https://youtu.be/Vq7kevsWlHk
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 +
~Interstigation
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==See also==
 +
{{AMC10 box|year=2023|ab=B|num-b=13|num-a=15}}
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 +
[[Category:Introductory Algebra Problems]]
 +
{{MAA Notice}}

Latest revision as of 22:05, 27 October 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

Solution 5 (Alternative Method for Manipulation)

$m^2 + mn + n^2 = m^2n^2$

$mn = m^2n^2 - m^2 - n^2$

$mn + 1 = m^2(n^2 - 1) - 1(n^2 - 1)$

$mn + 1 = (m + 1)(m - 1)(n + 1)(n - 1)$

Notice that the right side can be zero or one. If the right side is zero, m and n can be $(-1,1)$ and $(1,-1)$. If the right side is one, m and n can be $(0,0)$. There are $\boxed{\textbf{(C) 3}}$ solutions.

~unhappyfarmer

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

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