2020 AMC 10B Problem 9

1. The huge power looks scary, but the right side, $2y$, suggests completing the square in $y$. Move $2y$ to the left and add $1$ to both sides:
$$x^{2020} + y^2 - 2y + 1 = 1, \quad\text{that is}\quad x^{2020} + (y - 1)^2 = 1.$$
2. Both pieces on the left are whole numbers that are never negative. The power $2020$ is even, so $x^{2020}$ is never negative, and a square is never negative.
3. Two such numbers adding to $1$ must be $0$ and $1$ in some order. So there are two cases.
4. Case 1: $x^{2020} = 0$ and $(y - 1)^2 = 1$. Then $x = 0$, and $y - 1 = 1$ or $-1$, so $y = 2$ or $y = 0$. Pairs: $(0, 2)$ and $(0, 0)$.
5. Case 2: $x^{2020} = 1$ and $(y - 1)^2 = 0$. Then $y = 1$, and $x = 1$ or $x = -1$ (an even power of $-1$ is $1$). Pairs: $(1, 1)$ and $(-1, 1)$.
6. Any other integer $x$, such as $x = 2$, makes $x^{2020}$ far bigger than $1$, so there are no more solutions.

That makes $4$ ordered pairs.

The answer is $\boxed{\textbf{(D)}\ 4}$.

A big even exponent acts just like squaring for whole numbers: it gives $0$ at $x = 0$, $1$ at $x = 1$ and $x = -1$, and something huge for anything else. Completing the square turns the equation into "numbers that are never negative add up to something small", and that leaves only a few cases to check.

Move $2y$ to the left and complete the square in $y$:
$$x^{2020} + y^2 - 2y + 1 = 1 \quad\Longrightarrow\quad x^{2020} + (y-1)^2 = 1.$$
Both terms are nonnegative integers (the exponent $2020$ is even), and they add to $1$, so one term is $0$ and the other is $1$.

• $x^{2020} = 0$, $(y-1)^2 = 1$: $x = 0$ and $y = 0$ or $y = 2$. Two pairs.
• $x^{2020} = 1$, $(y-1)^2 = 0$: $x = \pm 1$ and $y = 1$. Two pairs.

That gives $(0,0), (0,2), (1,1), (-1,1)$, four ordered pairs.

The answer is $\boxed{\textbf{(D)}\ 4}$.