Diophantine equation in $\mathbb{Z_+}$

Question

Solve over positive integers: $$\left (1+\frac{1}{a} \right ) \left (1+\frac{1}{b} \right ) \left (1+\frac{1}{c} \right )=2$$

My claim is there aren't solutions, but I can't find a complete proof.

Answer 1

I think the following idea works.

Let $a=1+x$, $b=1+y$ and $c=1+z$.

Hence, we need to solve $$6+2(x+y+z)=xyz,$$ where $x$, $y$ and $z$ are non-negative integer numbers.

Thus, $$6+2(x+y)=z(xy-2)\geq xy-2,$$ which gives $$(x-2)(y-2)\leq12.$$

It's not so many cases.

Answer 2

Not a complete answer, but too long for a comment :

We can assume $1\le a\le b\le c$. For $c\le 1000$, there are the following solutions :

? for(a=1,1000,for(b=a,1000,for(c=b,1000,if((1+1/a)*(1+1/b)*(1+1/c)==2,print([a,
b,c])))))
[2, 4, 15]
[2, 5, 9]
[2, 6, 7]
[3, 3, 8]
[3, 4, 5]

No idea, whether further solutions exist. But it is clear that we cannot have $a\ge 4$ because then, we would have $$(1+\frac{1}{a})(1+\frac{1}{b})(1+\frac{1}{c})\le 1.25^3<2$$