Interesting Integer Equation

$\frac{1}{a}+\frac{1}{ab}+\frac{1}{abc}=1$ $a, b, c ∈ N$

$\frac{bc}{abc}+\frac{c}{abc}+\frac{1}{abc}=1$

$\frac{bc+c+1}{abc}=1$

$bc+c+1=abc$

$abc-bc-c=1$

$c(ab-b-1)=1$

In order for the equation to equal 1, $c$ must be equal to 1.

$\boxed{c=1}$

$ab-b-1=1$

$b(a-1)=2$

At this point, $b$ can either be 1, or 2.

$\boxed{b=1}$

$a-1=2$

$\boxed{a=3}$

and...

$\boxed{b=2}$

$2(a-1)=2$

$a-1=1$

$\boxed{a=2}$

This leaves us with a total of two solutions...

$\boxed{c=1, b= 1, a=3}$ and..

$\boxed{c=1, b= 2, a=2}$

If a, b, and c are all greater than 0, given the equation, we can deduce that 1/a >= 1/ab >=1/abc and their sum is equal to 1. Therefore, one outcome is when each factor is equal (resulting in 1/3 + 1/3 + 1/3) and one outcome when the first factor is larger than the other two. Since the second two factors in this case must be equal to yield a result of 1, there is only one such solution (1/2 + 1/4 + 1/4).

The equation can be re written as 1/c=b(a-1)-1 . clearly the rhs is an integer. For lhs to be an integer ..c must be equal to 1. Thus b(a-1)=2. The two possible solutions are 3,1,1 and 2,2,1

Given that $a,b,c$ are natrual numbers,

$\frac{1}{a}+\frac{1}{ab} +\frac{1}{abc}=1$

$\Leftrightarrow bc+c+1=abc$

$\Leftrightarrow 1=(ab-b-1)c$

$\Leftrightarrow c=1 \wedge 2= (a-1)b$

$\Leftrightarrow c=1 \wedge ((a-1=2 \wedge b=1) \vee (a-1=1 \wedge b=2))$

$\Leftrightarrow (a,b,c) \in \{ (3,1,1),(2,2,1) \}$

So there are $\boxed{2}$ ordered tuples

Since each term must get smaller, a can't equal any number 4 or greater (as 1/4+1/4+1/4 is less than 1)

So, we test by case.

a=1 is clearly impossible since it would require the other 2 terms to be 0.

a=2:

b=1 requires c to equal 0, so it doesn't work. b=2 works if c=1.

Since c will need to get smaller as b increases, and it equals 1 in the above scenario, there are no more solutions for a=2 besides the (2,1,1) we just found.

a=3:

b=1 works if c=1.

By the same logic as in a=2, we can see that (3,1,1) is the only solution.

We have now checked all cases, and found 2 solutions.