How many ordered pairs of integers (x,y) satisfy the equation: x^2020 +y^2=2y

Answer 1

Four solutions

x^2020 = 2y - y^2 = y (2-y) . If x is nonzero the left must be positive because the power of x is even

so y (2-y) must be positive , for an integer y

y y(2-y)

o. 0

1. 1

2. 0

if y<0 or y>2 then negative

so if x is nonzero, y has to be 1, giving x = 1 or -1

if x is zero then y is 0 or 2

so solutions

(1, 1) (-1,1) (0,0), (0,2)

note that 0 is technically an integer