Let the integers be x-1, x, x+1.
(x+1)² = (x-1)² + x² -140
x² + 2x + 1 = x² - 2x + 1 + x² - 140
2x = -2x + x² - 140
x² -4x -140 = 0
x² -14x +10x -140 = 0
x(x - 14) + 10(x - 14) = 0
(x - 14)(x + 10) = 0
x = 14 or -10
Going back to the criteria mentioned in the question and checking,
--- When x = -10
(-9)² = (-11)² + (-10)² - 140
81 = 121 + 100 - 140
81 = 81
--- When x = 14
(15)² = (13)² + (14)² - 140
225 = 169 + 196 - 140
225 = 225
Since both values satisfy the condition, both values are correct.
Hence, the 3 consecutive integers can be: -11, -10, -9 OR 13, 14, 15.
