2020 AMC 10B Problem 9

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 $\boxed{\textbf{(D)}\ 4}$.

A huge even exponent is a signal, not an obstacle: $x^{2020}$ behaves just like $x^2$ for integer solutions (it is $0$ at $x=0$, $1$ at $x = \pm1$, and enormous otherwise). Completing the square converts the equation into "nonnegative pieces summing to a tiny constant," which leaves only a handful of cases.

• Forgetting that x can be negative (x = -1 works because 2020 is even) and answering 3.
• Not completing the square and trying to factor $y^2 - 2y$ as $y(y-2)$, which does not bound $x$ cleanly.