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Difference between revisions of "2023 AMC 10B Problems/Problem 14"
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~Technodoggo
 
~Technodoggo
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==Solution==
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Case 1: <math>mn = 0</math>.
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 +
In this case, <math>m = n = 0</math>.
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Case 2: <math>mn \neq 0</math>.
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Denote <math>k = {\rm gcd} \left( m, n \right)</math>.
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Denote <math>m = k u</math> and <math>n = k v</math>.
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Thus, <math>{\rm gcd} \left( u, v \right) = 1</math>.
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Thus, the equation given in this problem can be written as
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<cmath>
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\[
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u^2 + uv + v^2 = k^2 u^2 v^2 .
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\]
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</cmath>
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Modulo <math>u</math>, we have <math>v^2 \equiv 0 \pmod{u}</math>.
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Because <math>\left( u, v \right) = 1</math>, we must have <math>|u| = |v| = 1</math>.
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Plugging this into the above equation, we get <math>2 + uv = k^2</math>.
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Thus, we must have <math>uv = -1</math> and <math>k = 1</math>.
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Thus, there are two solutions in this case: <math>\left( m , n \right) = \left( 1, -1 \right)</math> and <math>\left( m , n \right) = \left( -1, 1 \right)</math>.
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Putting all cases together, the total number of solutions is
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\boxed{\textbf{(C) 3}}.
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 +
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Revision as of 17:10, 15 November 2023

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

Solution

Obviously, $m=0,n=0$ is a solution.

\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

Solution

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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