Difference between revisions of "2023 AMC 10B Problems/Problem 14"
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== Solution 1 == | == Solution 1 == | ||
− | Clearly, <math>m=0,n=0</math> is | + | Clearly, <math>m=0,n=0</math> is one of the solutions. However, we can be quite sure that there are more, so we apply [https://artofproblemsolving.com/wiki/index.php/Simon%27s_Favorite_Factoring_Trick Simon's Favorite Factoring Trick] to get the following: |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
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m^2+mn+n^2 +mn &= m^2n^2 +mn\ | m^2+mn+n^2 +mn &= m^2n^2 +mn\ | ||
(m+n)^2 &= m^2n^2 +mn\ | (m+n)^2 &= m^2n^2 +mn\ | ||
− | (m+n)^2 &= mn(mn+1)\ | + | (m+n)^2 &= mn(mn+1).\ |
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | + | Essentially, this says that the product of two consecutive numbers <math>mn,mn+1</math> must be a perfect square. This 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=0</math> gives <math>(0,0)</math>. | ||
− | <math>mn=-1</math> gives <math>(1,-1), (-1,1)</math>. | + | <math>mn=-1</math> gives <math>(1,-1), (-1,1)</math>. Answer: <math>\boxed{\textbf{(C) 3}}.</math> |
~Technodoggo ~minor edits by lucaswujc | ~Technodoggo ~minor edits by lucaswujc | ||
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Modulo <math>u</math>, we have <math>v^2 \equiv 0 \pmod{u}</math>. | Modulo <math>u</math>, we have <math>v^2 \equiv 0 \pmod{u}</math>. | ||
− | Because <math>\left( u, v \right) = 1</math>, we must have <math>|u| = |v| = 1</math>. | + | Because <math>{\rm gcd} \left( u, v \right) = 1</math>., we must have <math>|u| = |v| = 1</math>. |
Plugging this into the above equation, we get <math>2 + uv = k^2</math>. | Plugging this into the above equation, we get <math>2 + uv = k^2</math>. | ||
Thus, we must have <math>uv = -1</math> and <math>k = 1</math>. | Thus, we must have <math>uv = -1</math> and <math>k = 1</math>. | ||
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
+ | ~ sravya_m18 | ||
==Solution 3 (Discriminant)== | ==Solution 3 (Discriminant)== | ||
− | 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>(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 | ||
+ | |||
+ | ==Solution 4 (Nice Substitution)== | ||
+ | Let <math>x=m+n, y=mn</math> then | ||
+ | <cmath>x^2-y=y^2</cmath> | ||
+ | |||
+ | Completing the square in <math>y</math> and multiplying by 4 then gives | ||
+ | <cmath>4x^2+1=(2y+1)^2</cmath> | ||
+ | |||
+ | Since the RHS is a square, clearly the only solutions are <math>x=0,y=0</math> and <math>x=0,y=-1</math>. | ||
+ | |||
+ | The first gives <math>(0,0)</math>. | ||
+ | |||
+ | The second gives <math>(-1,1)</math> and <math>(1,-1)</math> by solving it as a quadratic with roots <math>m</math> and <math>n</math>. | ||
+ | |||
+ | Thus there are <math>\boxed{\textbf{(C) 3}}</math> solutions. | ||
+ | |||
+ | ~ 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== | ||
https://youtu.be/5a5caco_YTo | https://youtu.be/5a5caco_YTo | ||
+ | ==Video Solution== | ||
+ | |||
+ | https://youtu.be/Dh1lDI1fHrw | ||
+ | |||
+ | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
+ | ==Video Solution by Interstigation== | ||
+ | |||
+ | https://youtu.be/Vq7kevsWlHk | ||
+ | |||
+ | ~Interstigation | ||
==See also== | ==See also== | ||
{{AMC10 box|year=2023|ab=B|num-b=13|num-a=15}} | {{AMC10 box|year=2023|ab=B|num-b=13|num-a=15}} | ||
+ | |||
+ | [[Category:Introductory Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest 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.