Let x = the 1st integer , then let x + 1 = the next integer
We can then write the following equation:
x(x+1) = [x + (x+1)] +1 and simplifying both sides gives
[tex] x^{2} + x = 2x + 2[/tex] Collecting all terms on the left side we have:
[tex] x^{2} +x-2x-2=0[/tex] simplify again
[tex] x^{2} -x-2=0[/tex] We can now solve by factoring:
(x - 2)(x + 1) = 0 or x = 2 and x = -1
Now if "X" can be 2 or -1 that means there are 2 possible sets of consecutive integers: 2,3 and -1,0
If you check, both of these satisfy our original conditions of the consecutive integers. Product = Sum + 1
