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−  ==Problem==
 +  #REDIRECT [[2020 AMC 10B Problems/Problem 9]] 
−   
−  How many ordered pairs of integers <math>(x, y)</math> satisfy the equation<cmath>x^{2020}+y^2=2y?</cmath>
 
−   
−  <math>\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } \text{infinitely many}</math>
 
−   
−  ==Solution, x first==
 
−  Set it up as a quadratic in terms of y:
 
−  <cmath>y^22y+x^{2020}=0</cmath>
 
−  Then the discriminant is
 
−  <cmath>\Delta = 44x^{2020}</cmath>
 
−  This will clearly only yield real solutions when <math>x^{2020} \leq 1</math>, because it is always positive.
 
−  Then <math>x=1,0,1</math>. Checking each one:
 
−  <math>1</math> and <math>1</math> are the same when raised to the 2020th power:
 
−  <cmath>y^22y+1=(y1)^2=0</cmath>
 
−  This has only has solutions <math>1</math>, so <math>(\pm 1,1)</math> are solutions.
 
−  Next, if <math>x=0</math>:
 
−  <cmath>y^22y=0</cmath>
 
−  Which has 2 solutions, so <math>(0,2)</math> and <math>(0,0)</math>
 
−   
−  These are the only 4 solutions, so <math>\boxed{\textbf{(D) } 4}</math>
 
−   
−  ==Solution 2, y first==
 
−   
−  Move the <math>y^2</math> term to the other side to get <math>x^{2020}=2yy^2 = y(2y)</math>. Because <math>x^{2020} \geq 0</math> for all <math>x</math>, then <math>y(2y) \geq 0 \Rightarrow y = 0,1,2</math>. If <math>y=0</math> or <math>y=2</math>, the right side is <math>0</math> and therefore <math>x=0</math>. When <math>y=1</math>, the right side become <math>1</math>, therefore <math>x=1,1</math>. Our solutions are <math>(0,2)</math>, <math>(0,0)</math>, <math>(1,1)</math>, <math>(1,1)</math>. There are <math>4</math> solutions, so the answer is <math>\boxed{\textbf{(D) } 4}</math>
 
−   
−  ==Video Solution==
 
−  https://youtu.be/6ujfjGLzVoE
 
−   
−  ~IceMatrix
 
−   
−  ==See Also==
 
−   
−  {{AMC12 boxyear=2020ab=Bnumb=7numa=9}}
 
−  {{MAA Notice}}
 