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

Revision as of 16:08, 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

@@ Line 1: / Line 1: @@
 How many ordered pairs of integers <math>(m, n)</math> satisfy the equation <math>m^2+mn+n^2 = m^2n^2</math>?
+== Solution ==
+Obviously, <math>m=0,n=0</math> is a solution.
+<cmath>\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*}</cmath>
+This basically say that the product of two consecutive numbers <math>mn,mn+1</math> must be a perfect square which 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>.
+~Technodoggo

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

Revision as of 16:08, 15 November 2023

Solution