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

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

@@ 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>.
 <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
@@ Line 54: / Line 54: @@
 ==Solution 3 (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>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
 ==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
+==Solution 5 (Alternative Method for Manipulation)==
+<math>m^2 + mn + n^2 = m^2n^2</math>
+<math>mn = m^2n^2 - m^2 - n^2</math>
+<math>mn + 1 = m^2(n^2 - 1) - 1(n^2 - 1)</math>
+<math>mn + 1 = (m + 1)(m - 1)(n + 1)(n - 1)</math>
+Notice that the right side can be zero or one.
+If the right side is zero, m and n can be <math>(-1,1)</math> and <math>(1,-1)</math>.
+If the right side is one, m and n can be <math>(0,0)</math>.
+There are <math>\boxed{\textbf{(C) 3}}</math> solutions.
+~unhappyfarmer
 ==Video Solution by OmegaLearn==
@@ Line 71: / Line 100: @@
 ~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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