Find equal sum: 1 + 2 + 3 = 1 * 2 * 3

[japiga]

Find all combinations of the three natural numbers which their sum is equal to their product of those factors (three natural numbers).
I have found just one combination 1 + 2 + 3 = 1 * 2 * 3. But it is only one combination, and the question is: is it only one combination and how to prove it?

[stapel_eliz]

...how to prove it?

Use algebra.

Assuming the numbers are meant to be integers or whole numbers, and assuming that they are meant to be consecutive, then you are trying to find solutions to x + (x + 1) + (x + 2) = x(x + 1)(x + 2).

[japiga]

Yes, and I got results (three solutions for x, x1 = 1, x2 = -1 i x3 = 3), and consequently, we have 3 consecutive numbers of each of x with following equations: 1 + 2 + 3 = 1 * 2 * 3; -1 + 0 + 1 = -1 * 0 * 1 and 3 + 4 + 5 is not equal to 3 * 4 * 5 (so in this case we my exclude solution for x3, or to put other numbers in opposite order, such as: 3, 2, 1 and now it gives a sense, 3 + 2 + 1 = 3 * 2 * 1. Do you think that I’ve solved it?

[stapel_eliz]

Yes, and I got results (three solutions for x, x1 = 1, x2 = -1 i x3 = 3)....

How did you arrive at these solutions? (If you plug them back into the original equation, provided earlier, which one(s) work?)

[japiga]

It works only if x1=1 and x2=-1, but not for x3=3. Is it now correct answer on it?
I arrived on it just solving the equation: x + (x + 1) + (x + 2) = x(x + 1)(x + 2)
x + x + 1 + x + 2 = x(x2 + 2x + x +2)
3x + 3 = x3 + 3x2 + 2x
x3 - 3x2 – x + 3 = 0
x2 (x- 3) – (x – 3) = 0
(x2 – 1)(x -3) = 0

[stapel_eliz]

3x + 3 = x3 + 3x2 + 2x
x3 - 3x2 – x + 3 = 0

You were correct to here:

. . . . .

You subtracted the and the from the left-hand side to the right-hand side:

. . . . .

Then you simplified. But how did you end up with your second line in the quote above?

[japiga]

II have just solved it! We have 3 solutions for x:
x1=1; x2=-1; and x3=-3. So, we have same result when we summarize and multiply all factors x, (x+1) and (x+2).