Answer:
3
Step-by-step explanation:
since the 3 integers are consecutive, we are dealing with x, x+1, x+2.
and their sum is the same as their product :
x + (x + 1) + (x + 2) = x(x + 1)(x + 2)
3x + 3 = x(x² + 3x + 2) = x³ + 3x² + 2x
x³ + 3x² - x - 3 = 0
this is a polynomial of third degree.
and as such it has 3 solutions.
of course, it could be that some of them are the same or are even in the realm of complex numbers (i = sqrt(-1)), but usually these 3 solutions are different real numbers.
I tried x=1 just to see, and, hey, it is a solution for this equation.
x = 1 means that the other 2 consecutive integers are 2 and 3.
and indeed, 1+2+3 = 1×2×3 = 6.
now it is easier to find the other 2 solutions, as a zero solution can be expressed as a factor of the whole expression.
for x = 1 the factor term is (x - 1), as this term is then turning 0, when x = 1.
I can divide the main expression by this factor and then analyze the quotient about the other 2 solutions.
x³ + 3x² - x - 3 : x - 1 = x² + 4x + 3
- x³ - x²
----------------
0 4x² - x
- 4x² - 4x
-----------------------
0 + 3x - 3
- 3x - 3
---------------------------
0 0
so, the original expression can be written as
(x² + 4x + 3)(x - 1).
now we need to find the 2 zero solutions for x²+4x+3
the general solution to a quadratic equation is
x = (-b ± sqrt(b² - 4ac))/(2a)
in our case
a = 1
b = 4
c = 3
so,
x = (-4 ± sqrt(4² - 4×1×3))/(2×1) =
= (-4 ± sqrt(16 - 12))/2 = (-4 ± sqrt(4))/2 =
= (-4 ± 2)/2 = -2 ± 1
x1 = -2 + 1 = -1
x2 = -2 - 1 = -3
so, we have the additional solutions :
-1 0 1
-3 -2 -1
-1 + 0 + 1 = -1×0×1 = 0
-3 + -2 + -1 = -3×-2×-1 = -6
and there we have it fully proven :
there are 3 different sets of 3 consecutive integers with the same sum as product.
