Difference between revisions of "2023 AMC 10B Problems/Problem 14"
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==Problem== | ==Problem== | ||
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>? | ||
+ | |||
+ | <math>\textbf{(A) }7\qquad\textbf{(B) }1\qquad\textbf{(C) }3\qquad\textbf{(D) }6\qquad\textbf{(E) }5</math> | ||
== Solution 1 == | == Solution 1 == |
Revision as of 20:41, 15 November 2023
Contents
[hide]Problem
How many ordered pairs of integers satisfy the equation ?
Solution 1
Clearly, is 1 solution. However there are definitely more, so we apply Simon's Favorite Factoring Expression to get this:
This basically say that the product of two consecutive numbers must be a perfect square which is practically impossible except or . gives . gives .
~Technodoggo ~minor edits by lucaswujc
Solution 2
Case 1: .
In this case, .
Case 2: .
Denote . Denote and . Thus, .
Thus, the equation given in this problem can be written as
Modulo , we have . Because , we must have . Plugging this into the above equation, we get . Thus, we must have and .
Thus, there are two solutions in this case: and .
Putting all cases together, the total number of solutions is .
~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 to get The discriminant of this quadratic is For to be an integer, we must have be a perfect square. Thus, either is a perfect square or and . The first case gives , which result in the equations and , for a total of two pairs: and . The second case gives the equation , so it's only pair is . In total, the total number of solutions is .
~A_MatheMagician
Video Solution by OmegaLearn
See also
2023 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.