Question 757854
consecutive positive integers: x, (x+1), (x+2), (x+3), (x+4)


2(x)(x+4) = (x+2)^2 + 41
2(x^2 + 4x) = x^2 + 4x + 4 + 41
2x^2 + 8x = x^2 + 4x + 45
x^2 + 4x - 45 = 0
(x + 9)(x - 5) = 0
x = -9, x = 5


x has to be positive, so the middle integer is x + 2 = 7.