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

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Adding by \( 2mn \) gives us \( (m+n)^2 = m^2n^2 + mn \).
 
Adding by \( 2mn \) gives us \( (m+n)^2 = m^2n^2 + mn \).
  
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We aren't done yet though! \( m^2n^2 + mn \) can be simplified int \( mn(mn+1) \), giving us \( (m+n)^2 = mn(mn+1) \).  
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We aren't done yet though! \( m^2n^2 + mn \) can be simplified into \( mn(mn+1) \), giving us \( (m+n)^2 = mn(mn+1) \).  
  
 
Okay so now we're done, but, Pinotation, this doesn't do anything! Well, now we can use the Zero Product Property!  
 
Okay so now we're done, but, Pinotation, this doesn't do anything! Well, now we can use the Zero Product Property!  
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\end{align*}</cmath>
 
\end{align*}</cmath>
  
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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>.  
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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 impossible except for when <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>
  
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~Technodoggo ~minor edits by lucaswujc
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~Technodoggo ~minor edits by lucaswujc ~minor edits by luke22
  
 
==Solution 3 ==
 
==Solution 3 ==
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~ sravya_m18
 
~ sravya_m18
  
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==Solution 3 (Discriminant)==
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==Solution 4 (Discriminant)==
 
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>.
 
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
  
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==Solution 4 (Nice Substitution)==
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==Solution 5 (Nice Substitution)==
 
Let <math>x=m+n, y=mn</math> then  
 
Let <math>x=m+n, y=mn</math> then  
 
<cmath>x^2-y=y^2</cmath>
 
<cmath>x^2-y=y^2</cmath>
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~ Grolarbear
 
~ Grolarbear
  
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==Solution 5 (Alternative Method for Manipulation)==
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==Solution 6 (Alternative Method for Manipulation)==
  
 
<math>m^2 + mn + n^2 = m^2n^2</math>
 
<math>m^2 + mn + n^2 = m^2n^2</math>
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~unhappyfarmer
 
~unhappyfarmer
  
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==Solution 6 (Obtaining ranges)==
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==Solution 7 (Obtaining ranges)==
  
 
Set <math>m\leq n</math>. Then, we can say that  
 
Set <math>m\leq n</math>. Then, we can say that  
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-lisztepos
 
-lisztepos
  
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==Solution 7 (Inequality)==
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==Solution 8 (Inequality)==
  
 
If <math>mn = 0</math>, then <math>m^2 + 0 + n^2 = 0</math>, <math>(m, n) = (0, 0)</math>.
 
If <math>mn = 0</math>, then <math>m^2 + 0 + n^2 = 0</math>, <math>(m, n) = (0, 0)</math>.

Latest revision as of 22:00, 13 September 2025

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

Let's use 10th grade math to solve this. After all, it is called the AMC 10 for a reason!

We have \[m^2 + mn + n^2 = m^2n^2.\] We subtract \( mn \) on both sides to get

\( m^2 + n^2 = m^2n^2 - mn \).

Fun Fact! You can write \( m^2 + n^2 \) as \( (m+n)^2 - 2mn \)! Let's use this!

We convert the Left Hand Side into \( (m+n)^2 - 2mn \) to get \( (m+n)^2 - 2mn = m^2n^2 - mn \).

Adding by \( 2mn \) gives us \( (m+n)^2 = m^2n^2 + mn \).

We aren't done yet though! \( m^2n^2 + mn \) can be simplified into \( mn(mn+1) \), giving us \( (m+n)^2 = mn(mn+1) \).

Okay so now we're done, but, Pinotation, this doesn't do anything! Well, now we can use the Zero Product Property!

How? I'll show you!

We subtract \( mn(mn+1) \) on both sides of the equation to get \( (m+n)^2 - mn(mn+1) = 0 \). Now we do a bit of casework.

Notice how \( (m+n)^2 - mn(mn+1) = 0 \) is just \( (m+n)(m+n) - mn(mn+1) = 0 \). So, either \( m+n = 0 \) and \( mn = 0 \), or \( m +n = 0 \) and \( mn+1 = 0 \). Let's look at it through both cases.

Case 1: \( m+n = 0 \) and \( mn = 0 \). If \( m+n = 0 \) and \( mn = 0 \), then that must mean that either \( m = 0 \) or \( n = 0 \), and if we substitute either \( m=0 \) or \( n=0 \) in, we still get either \( m=0 \) or \( n=0 \), so therefore we have 1 ordered pair, \( (0,0) \).

Case 2: \( m +n = 0 \) and \( mn+1 = 0 \). \( mn+1 =0 \) means that \( mn=-1 \). So, for this to be possible, either \( m = -1 \) and \( n=1 \), or \( m=1 \) and \( n=-1 \). Let's check for contradictions quickly. We see that \( m + n = 0 \), and \( -1 + 1 = 0 \) and \( 1 - 1 =0 \), so we know the ordered pairs \( (-1,1) \) and \( (1,-1) \) both work.

We have a total of $\boxed{\textbf{(C) 3}}.$ ordered pairs. \( (-1,1) \), \( (0,0) \), and \( (1,-1) \).

~Pinotation

Solution 2

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 impossible except for when $mn=0$ 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 ~minor edits by luke22

Solution 3

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 4 (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 5 (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 6 (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

Solution 7 (Obtaining ranges)

Set $m\leq n$. Then, we can say that

\[3n^2\geq m^2n^2\]

\[3\geq m^2\]

Or $-\sqrt{3} \leq m \leq \sqrt{3}$, and since we are dealing with integers, $m=-1$, $0$ or $1$. Testing these numbers, we get $n=1$, $n=0$ and $n=-1$ respectively. Although the solution $(1,-1)$ is a solution in the end, our initial condition for this case was $m\leq n$. For better rigour, we can just consider the other case $m>n$ to validate solution $(1,-1)$.

-lisztepos

Solution 8 (Inequality)

If $mn = 0$, then $m^2 + 0 + n^2 = 0$, $(m, n) = (0, 0)$.

If $mn \neq 0$, then

\[\frac{m^2 + mn + n^2}{m^2n^2} = 1\]

\[\frac{1}{m^2} + \frac{1}{mn} + \frac{1}{n^2} = 1\]

\[\frac{1}{m^2} + \frac{2}{|mn|} + \frac{1}{n^2} > 1\]

\[\left(\frac{1}{|m|} + \frac{1}{|n|}\right)^2 > 1\]

\[\frac{1}{|m|} + \frac{1}{|n|} > 1\]

Obviously, at least one of $|m|, |n|$ is 1. If $m = 1$, $1 + n + n^2 = n^2 \Rightarrow n = -1$. If $m = -1$, $1 - n + n^2 = n^2 \Rightarrow n = 1$. We omit the discussion of $n = \pm 1$.

Finally, we get $(m, n) = (0, 0), (1, -1), (-1, 1)$, there are $\boxed{\textbf{(C) 3}}$ solutions.

~reda_mandymath

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

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

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