2023 AMC 10B Problems/Problem 14: Difference between revisions

Revision as of 03:39, 17 July 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

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

Solution 6 (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 7 (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

@@ Line 103: / Line 103: @@
 -lisztepos
+==Solution 7 (Inequality)==
+If <math>mn = 0</math>, then <math>m^2 + 0 + n^2 = 0</math>, <math>(m, n) = (0, 0)</math>.
+If <math>mn \neq 0</math>, then
+<cmath>\frac{m^2 + mn + n^2}{m^2n^2} = 1</cmath>
+<cmath>\frac{1}{m^2} + \frac{1}{mn} + \frac{1}{n^2} = 1</cmath>
+<cmath>\frac{1}{m^2} + \frac{2}{|mn|} + \frac{1}{n^2} > 1</cmath>
+<cmath>\left(\frac{1}{|m|} + \frac{1}{|n|}\right)^2 > 1</cmath>
+<cmath>\frac{1}{|m|} + \frac{1}{|n|} > 1</cmath>
+Obviously, at least one of <math>|m|, |n|</math> is 1. If <math>m = 1</math>, <math>1 + n + n^2 = n^2 \Rightarrow n = -1</math>. If <math>m = -1</math>, <math>1 - n + n^2 = n^2 \Rightarrow n = 1</math>. We omit the discussion of <math>n = \pm 1</math>.
+Finally, we get <math>(m, n) = (0, 0), (1, -1), (-1, 1)</math>, there are <math>\boxed{\textbf{(C) 3}}</math> solutions.
+~[https://artofproblemsolving.com/wiki/index.php/User:Reda_mandymath reda_mandymath]
 ==Video Solution by OmegaLearn==

2023 AMC 10B (Problems • Answer Key • Resources)
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