Difference between revisions of "2020 AMC 12B Problems/Problem 8"

(Solution)
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Then the discriminant is
 
Then the discriminant is
 
<cmath>\Delta = 4-4x^{2020}</cmath>
 
<cmath>\Delta = 4-4x^{2020}</cmath>
This will clearly only yield real solutions when <math>x^{2020} \leq 1</math>
+
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>
  
 
==See Also==
 
==See Also==

Revision as of 19:28, 7 February 2020

Problem

How many ordered pairs of integers $(x,y)$ satisfy the equation \[x^{2020}+y^2=2y\]

Solution

Set it up as a quadratic in terms of y: \[y^2-2y+x^{2020}=0\] Then the discriminant is \[\Delta = 4-4x^{2020}\] This will clearly only yield real solutions when $x^{2020} \leq 1$, because it is always positive. Then $x=-1,0,1$. Checking each one: $-1$ and $1$ are the same when raised to the 2020th power: \[y^2-2y+1=(y-1)^2=0\] This has only has solutions $1$, so $(\pm 1,1)$ are solutions. Next, if $x=0$: \[y^2-2y=0\] Which has 2 solutions, so $(0,2)$ and $(0,0)$

These are the only 4 solutions, so $\boxed{D}$

See Also

2020 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
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 12 Problems and Solutions

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