2020 AMC 10B Problems/Problem 9

The following problem is from both the 2020 AMC 10B #9 and 2020 AMC 12B #8, so both problems redirect to this page.

Problem

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

$\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } \text{infinitely many}$

Solutions

Solution 1

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: Solve for 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 the discriminant must be 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 $x=0$ : $y^2-2y=0 \Rightarrow y(y-2)=0$ Which has 2 solutions, so $(0,2)$ and $(0,0)$ .

These are the only 4 solutions, so our answer is $\boxed{\textbf{(D) } 4}$ .

~edits by BakedPotato66

Solution 4: Solve for 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 $(0,2)$ , $(0,0)$ , $(1,1)$ , $(-1,1)$ . These are the only $4$ solutions, so the answer is $\boxed{\textbf{(D) } 4}$

- wwt7535

~ edits by BakedPotato66

Solution 5: Similar to solution 4

Since $x^{2020}$ and $y^2$ are perfect squares, they are both nonnegative. That means $y^2$ plus a nonnegative number equals $2y$ , which means $y^2 \leq 2y.$ The only possible integer values for $y$ are $0, 1, 2$ .

For $y=0$ , $x$ can only be $0$ .

For $y=1$ , $x^2=1$ so $x=1, -1$ .

For $y=2$ , $x$ can only be $0$ as well.

This gives us the solutions $(0, 0)$ , $(1, 1)$ , $(-1, 1)$ , and $(0, 2)$ . These are the only solutions, so there is a total of $\boxed{\textbf{(D) } 4}$ ordered pairs.

- kc1374

Solution 6: (Casework)

We see that $x$ has to be $1$ , $0$ , or $-1$ , as any other integer would make this value too large. We also know that because $2020$ is even, both $-1$ , and $1$ for $x$ will yield the same $x^{2020}$ value of $1$ .

Case 1

$x=0$

$y^2 = 2y$

$y$

$y = 2$

$y = 0$

Case 2

$x = 1$

$-1$

$1 + y^2 = 2y$

$2y$

$y^2-2y + 1 = 0$

$(y-1)^2 = 0$

$y = 1$

$x$

$-1$

$1$

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

Solution 7: (Casework)

We can move the $y^2$ term to the other side to obtain $x^{2020} = 2y - y^2$ , which can be easily factored into $x^{2020} = (2-y)y$ . If we raise each side to the power 1/2020, we get $x = ((2-y)(y))^{1/2020}$ . Since we're dealing with integer values, the inside of the square root and thus $(2-y)y$ must be greater than $0$ . For this to happen, $y$ must be between $0$ and $2$ inclusive. Testing each integer value, we get the cases:

Case 1( $y = 0$ ): When $y=0$ , then $y(2-y)$ is also 0, and since the square root of $0$ is $0$ , $x$ must also be $0$ . So our first pair of solutions is $(0,0)$ .

Case 2( $y = 1$ ): When $y=1$ , then $y(2-y)$ is also 1. The possible values for $x$ are $1$ and $-1$ , so we get 2 more pairs of solutions: $(1, -1)$ and $(1,1)$ .

Case 3( $y = 2$ ): When $y=2$ , then $y(2-y)$ is $0$ , and since the square root of $0$ is $0$ , $x$ must also be $0$ - so we get one more pair of solutions, $(2,0)$

If we count all these pairs( $(0,0),(1,-1),(1,1),(2,0)$ ), we have 4 pairs of solutions, so our answer is $\boxed{\textbf{(D) } 4}$ ordered pairs.

- abed_nadir

Video Solution (HOW TO CREATIVELY PROBLEM SOLVE!!!)

https://youtu.be/FfATkdxncG4

~Education, the Study of Everything

Video Solution 1 by TheBeautyOfMath

https://youtu.be/6ujfjGLzVoE

Video Solution by WhyMath

https://youtu.be/7dQ423hhgac

~savannahsolver

Video Solution by Sohil Rathi

https://youtu.be/zfChnbMGLVQ?t=4251

2020 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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 10 Problems and Solutions

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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