2020 AMC 12B Problems/Problem 8

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 $\boxed{D}$

Solution 2

Move the $y^2$ term to the other side to get $x^{2020}=2y-y^2 = y(2-y)$ . Because $x^{2020} \geq 0$ for all $x$ , then $y(2-y) \geq 0 \Rightarrow y = 0,1,2$ . If $y=0$ or $y=2$ , the right side is $0$ and therefore $x=0$ . When $y=1$ , the right side become $1$ , therefore $x=1,-1$ . Our solutions are $(0,2)$ , $(0,0)$ , $(1,1)$ , $(-1,1)$ . There are $4$ solutions, so the answer is $\boxed{D}$

2020 AMC 12B (Problems • Answer Key • Resources)
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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2020 AMC 12B Problems/Problem 8

Contents

Problem

Solution

Solution 2

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