AMC12 2020 B

AMC12 2020 B · Q8

AMC12 2020 B · Q8. It mainly tests Absolute value, Perfect squares & cubes.

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

有整数对 $(x, y)$ 多少对满足方程 $x^{2020} + y^2 = 2y$？

(A) 1 1

(B) 2 2

(D) 4 4

(E) infinitely many 无穷多个

Answer

Correct choice: (D)

正确答案：（D）

Solution

The given equation is equivalent to $$x^{2020} + y^2 - 2y + 1 = 1,$$ which is equivalent to $x^{2020} + (y - 1)^2 = 1$. The only way to write 1 as the sum of the squares of two integers is as $0 + 1$ or $1 + 0$, so $x$ must be $0$ or $\pm 1$. The complete set of solutions is $\{(0,0), (0,2), (1,1), (-1,1)\}$, which has 4 elements.

给定方程等价于 $$x^{2020} + y^2 - 2y + 1 = 1,$$ 即 $x^{2020} + (y - 1)^2 = 1$。1 作为两个整数平方和的唯一方式是 $0 + 1$ 或 $1 + 0$，故 $x$ 必为 $0$ 或 $\pm 1$。所有解为 $\{(0,0), (0,2), (1,1), (-1,1)\}$，共 4 个。

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