Difference between revisions of "2023 AMC 10B Problems/Problem 14"
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− | We can move all terms to one side and | + | 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>4m^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>, 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 14:14, 29 March 2024
Contents
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
, 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)
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.