find four consecutive integers such that the sum of the squares of the first two is 11 less than the square of the fourth

Answer 1

Let x, x+1, x+2, x+3 be the four consecutive integers.

"the sum of the squares of the first two is 11 less than the square of the fourth"
x^2 + (x+1)^2 + 11 = (x+3)^2
x^2 + x^2 + 2x + 1 + 11 = x^2 + 6x + 9
2x^2 + 2x + 12 = x^2 + 6x + 9
x^2 - 4x + 3 = 0
(x-3)(x-1) = 0
x = 1, 3

So the four consecutive integers are 1, 2, 3, 4; or 3, 4, 5, 6.