Difference between revisions of "2023 AMC 10B Problems/Problem 14"
m (→Solution 4 (Nice Substitution)) |
(→Solution 4 (Nice Substitution)) |
||
Line 74: | Line 74: | ||
~ Grolarbear | ~ Grolarbear | ||
+ | |||
+ | ==Solution 5 (Alternative Method for Manipulation)== | ||
+ | |||
+ | <math>m^2 + mn + n^2 = m^2n^2</math> | ||
+ | <math>mn = m^2n^2 - m^2 - n^2</math> | ||
+ | <math>mn + 1 = m^2(n^2 - 1) - 1(n^2 - 1)</math> | ||
+ | <math>mn + 1 = (m + 1)(m - 1)(n + 1)(n - 1)</math> | ||
+ | Notice that the right side can be zero or one. | ||
+ | If the right side is zero, m and n can be <math>(-1,1)</math> and <math>(1,-1)</math>. | ||
+ | If the right side is one, m and n can be <math>(0,0)</math>. | ||
+ | There are <math>\boxed{\textbf{(C) 3}}</math> solutions. | ||
+ | |||
+ | ~unhappyfarmer | ||
==Video Solution by OmegaLearn== | ==Video Solution by OmegaLearn== |
Revision as of 22:05, 27 October 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 in and multiplying by 4 then gives
Since the RHS is a square, clearly the only solutions are and .
The first gives .
The second gives and by solving it as a quadratic with roots and .
Thus there are solutions.
~ Grolarbear
Solution 5 (Alternative Method for Manipulation)
Notice that the right side can be zero or one. If the right side is zero, m and n can be and . If the right side is one, m and n can be . There are solutions.
~unhappyfarmer
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.