Question 157974
    Find 2 consecutive odd integers whose product is one less than 6 times thier sum. Doman for the smallest {-1, 1, 11}
.
Let x = 1st consecutive odd integer
then
x+2 = 2nd consecutive odd integer
.
From:"whose product is one less than 6 times their sum"
x(x+2) = 6(x+(x+2))-1
x(x+2) = 6(2x+2)-1
x^2+2x = 12x+12-1
x^2+2x = 12x+11
x^2-10x = 11
x^2-10x-11 = 0
.
Factoring the left:
(x-11)(x+1)=0
x = {-1,11}
.
Two possible sets:
-1 and 1
11 and 13