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

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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>?


== Solution ==
== Solution 1 ==


Obviously, <math>m=0,n=0</math> is a solution.
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:


<cmath>\begin{align*}
<cmath>\begin{align*}
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<math>mn=-1</math> gives <math>(1,-1), (-1,1)</math>.
<math>mn=-1</math> gives <math>(1,-1), (-1,1)</math>.


~Technodoggo
~Technodoggo ~minor edits by lucaswujc


==Solution==
==Solution 2 ==


Case 1: <math>mn = 0</math>.
Case 1: <math>mn = 0</math>.

Revision as of 18:15, 15 November 2023

How many ordered pairs of integers $(m, n)$ satisfy the equation $m^2+mn+n^2 = m^2n^2$?

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)

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