Having trouble with this NRP

There are only two integer triples that satisfy the equation ab+ac+bc=a+b+c+abc. They are:

(1, 2, 5) (2, 3, 4)

To prove this, we can first subtract abc from both sides of the equation to get:

ab+ac+bc=a+b+c

Now, we can factor the left-hand side of the equation as:

(a+b)(c-1)=a+b+c

We can see that this equation will only have integer solutions if (a+b) and (c-1) are both factors of (a+b+c).

The only positive integer pairs (a+b) and (c-1) that are factors of (a+b+c) are (1, 5) and (2, 4).

This means that the only possible solutions for a, b, and c are a=1, b=2, c=5 and a=2, b=3, c=4.

I hope this helps! Let me know if you have any other questions.

Here are 4 integer triplets that satisfy the equation:

         a   b   c

1  = (0, 0, 0)
2  = (2, 2, 0)
3  = (2, 0, 2)
4  = (0, 2, 2)