# Difference between revisions of "2020 AMC 10B Problems/Problem 9"

## Problem

How many ordered pairs of integers $(x, y)$ satisfy the equation $$x^{2020}+y^2=2y?$$ $\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } \text{infinitely many}$

## Solution

Rearranging the terms and and completing the square for $y$ yields the result $x^{2020}+(y-1)^2=1$. Then, notice that $x$ can only be $0$, $1$ and $-1$ because any value of $x^{2020}$ that is greater than 1 will cause the term $(y-1)^2$ to be less than $0$, which is impossible as $y$ must be real. Therefore, plugging in the above values for $x$ gives the ordered pairs $(0,0)$, $(1,1)$, $(-1,1)$, and $(0,2)$ gives a total of $\boxed{\textbf{(D) }4}$ ordered pairs.

## Solution 2

Bringing all of the terms to the LHS, we see a quadratic equation $$y^2 - 2y + x^{2020} = 0$$ in terms of $y$. Applying the quadratic formula, we get $$y = \frac{2\pm\sqrt{4-4\cdot1\cdot x^{2020}}}{2}=\frac{2\pm\sqrt{4(1-x^{2020})}}{2}.$$ In order for $y$ to be real, which it must be given the stipulation that we are seeking integral answers, we know that the discriminant, $4(1-x^{2020})$ must be nonnegative. Therefore, $$4(1-x^{2020}) \geq 0 \implies x^{2020} \leq 1.$$ Here, we see that we must split the inequality into a compound, resulting in $-1 \leq x \leq 1$.

The only integers that satisfy this are $x \in \{-1,0,1\}$. Plugging these values back into the quadratic equation, we see that $x = \{-1,1\}$ both produce a discriminant of $0$, meaning that there is only 1 solution for $y$. If $x = \{0\}$, then the discriminant is nonzero, therefore resulting in two solutions for $y$.

Thus, the answer is $2 \cdot 1 + 1 \cdot 2 = \boxed{\textbf{(D) }4}$.

~Tiblis

## Solution 3, x first

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

## Solution 4, y first

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

~IceMatrix