The square of the largest of 3 consecutive integers is 140 less than the sum of the squares of the two smaller integers. find the 3 integers

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.