What is the number of ordered pair of integers (x,y) satisfying the equation x^2 + 6x + y^2 = 4?

1 Answer
Feb 23, 2018

88

Explanation:

"Complete the square for x : "Complete the square for x :

"(x+3)^2 + y^2 = 13(x+3)2+y2=13

"As both terms are positive, we know that"As both terms are positive, we know that

-4 < x+3 < 44<x+3<4
"and"and
-4 < y < 44<y<4

y = pm 3 => x+3 = pm 2 => x = -5 or -1y=±3x+3=±2x=5or1
y = pm 2 => x+3 = pm 3 => x = -6 or 0y=±2x+3=±3x=6or0
y = pm 1 " and " y = 0, " yield no perfect square"y=±1 and y=0, yield no perfect square

"So we have 8 solutions : "So we have 8 solutions :
(-5, -3), (-5, 3), (-1, -3), (-1, 3),(5,3),(5,3),(1,3),(1,3),
(-6, -2), (-6, 2), (0, -2), (0, 2).(6,2),(6,2),(0,2),(0,2).