Difference between revisions of "2023 AMC 10B Problems/Problem 14"
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− | We can move all terms to one side and write the equation as a quadratic in terms of <math>n</math> to get <cmath>(1-m^2)n^2+(m)n+(m^2)=0.</cmath> The discriminant of this quadratic is <cmath>\Delta = m^2-4(1-m^2)(m^2)=m^2(4m^2-3).</cmath> For <math>n</math> to be an integer, we must have <math>m^2(4m^2-3)</math> be a perfect square. Thus, either <math> | + | We can move all terms to one side and write the equation as a quadratic in terms of <math>n</math> to get <cmath>(1-m^2)n^2+(m)n+(m^2)=0.</cmath> The discriminant of this quadratic is <cmath>\Delta = m^2-4(1-m^2)(m^2)=m^2(4m^2-3).</cmath> For <math>n</math> to be an integer, we must have <math>m^2(4m^2-3)</math> be a perfect square. Thus, either <math>(2m)^2-3</math> is a perfect square or <math>m^2 = 0</math> and <math>m = 0</math>. The first case gives <math>m=-1,1</math> (larger squares are separated by more than 3), which result in the equations <math>-n+1=0</math> and <math>n-1=0</math>, for a total of two pairs: <math>(-1,1)</math> and <math>(1,-1)</math>. The second case gives the equation <math>n^2=0</math>, so it's only pair is <math>(0,0)</math>. In total, the total number of solutions is <math>\boxed{\textbf{(C) 3}}</math>. |
~A_MatheMagician | ~A_MatheMagician |
Revision as of 21:06, 9 July 2024
Contents
[hide]Problem
How many ordered pairs of integers satisfy the equation ?
Solution 1
Clearly, is one of the solutions. However, we can be quite sure that there are more, so we apply Simon's Favorite Factoring Trick to get the following:
Essentially, this says that the product of two consecutive numbers must be a perfect square. This is practically impossible except or . gives . gives . Answer:
~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) ~ sravya_m18
Solution 3 (Discriminant)
We can move all terms to one side and write 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 (larger squares are separated by more than 3), 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
Solution 4 (Nice Substitution)
Let then Completing the square then gives Since the RHS is a square, clearly the only solutions are and . The first gives while the second gives and by solving it as a quadratic with roots and . Thus there are solutions.
~ Grolarbear
Video Solution by OmegaLearn
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution by Interstigation
~Interstigation
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.