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| ==Problem==
| | #REDIRECT [[2020 AMC 10B Problems/Problem 9]] |
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| How many ordered pairs of integers <math>(x,y)</math> satisfy the equation <cmath>x^{2020}+y^2=2y</cmath>
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| ==Solution==
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| Set it up as a quadratic in terms of y:
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| <cmath>y^2-2y+x^{2020}=0</cmath>
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| Then the discriminant is
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| <cmath>\Delta = 4-4x^{2020}</cmath>
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| This will clearly only yield real solutions when <math>x^{2020} \leq 1</math>, because it is always positive.
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| Then <math>x=-1,0,1</math>. Checking each one:
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| <math>-1</math> and <math>1</math> are the same when raised to the 2020th power:
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| <cmath>y^2-2y+1=(y-1)^2=0</cmath>
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| This has only has solutions <math>1</math>, so <math>(\pm 1,1)</math> are solutions.
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| Next, if <math>x=0</math>:
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| <cmath>y^2-2y=0</cmath>
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| Which has 2 solutions, so <math>(0,2)</math> and <math>(0,0)</math>
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| These are the only 4 solutions, so <math>\boxed{D}</math>
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| ==Solution 2==
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| Move the <math>y^2</math> term to the other side to get <math>x^{2020}=2y-y^2 = y(2-y)</math>. Because <math>x^{2020} \geq 0</math> for all <math>x</math>, then <math>y(2-y) \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{D}</math>
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| ==See Also==
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| {{AMC12 box|year=2020|ab=B|num-b=7|num-a=9}}
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| {{MAA Notice}}
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