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

Revision as of 14:45, 25 November 2023

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 1 solution. However there are definitely more, so we apply Simon's Favorite Factoring Expression to get this:

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

This basically say that the product of two consecutive numbers $mn,mn+1$ must be a perfect square which is practically impossible except $mn=0$ or $mn+1=0$ . $mn=0$ gives $(0,0)$ . $mn=-1$ gives $(1,-1), (-1,1)$ .

~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 $\left( u, v \right) = 1$ , 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)

Solution 3 (Discriminant)

We can move all terms to one side and wrote 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 $4m^2-3$ is a perfect square or $m^2 = 0$ and $m = 0$ . The first case gives $m=-1,1$ , 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 $x=m+n, y=xn$ then $x^2-y=y^2$ Completing the square 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 $(0,0)$ while 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)

@@ Line 56: / Line 56: @@
 ~A_MatheMagician
+==Solution 4 (Nice Substitution)==
+Let <math>x=m+n, y=xn</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.
+~ Grolarbear
 ==Video Solution by OmegaLearn==

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