Find all distinct positive integers $a,b,c,d,e$ with some given conditions. Find all distinct positive integers $a,b,c,d,e$ such that
$$\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)\left(1+\frac{1}{d}\right)\left(1+\frac{1}{e}\right)-\frac{1}{abcde}$$    
is a positive integer.  None of $a,b,c,d,e$ is $1$.
Is there any method for this? I have no idea. I can just fix the limit. I found that if there exist such integers, the expression can be $1,2$ or $3$.
 A: Consider 
$$2\le a \le b \le c \le d \le e.$$
Denote
$$F(a)=\dfrac{a+1}{a},$$
$$F(a,b)=\dfrac{a+1}{a}\cdot \dfrac{b+1}{b},$$
$$...$$
$$F(a,b,c,d,e)=\dfrac{a+1}{a}\cdot \dfrac{b+1}{b}\cdot \dfrac{c+1}{c}\cdot \dfrac{d+1}{d}\cdot \dfrac{e+1}{e}.
$$
Note, that
$$
(a+1)(b+1)(c+1)(d+1)(e+1)>abcde+1
$$
$$
\Downarrow
$$
$$
F(a,b,c,d,e)-\dfrac{1}{abcde}>1\tag{1}.
$$
Note, that
$$
F(a,b)-\dfrac{1}{ab}>F(a),\\
F(a,b,c)-\dfrac{1}{abc}>F(a,b),\\
F(a,b,c,d)-\dfrac{1}{abcd}>F(a,b,c),\\
F(a,b,c,d,e)-\dfrac{1}{abcde}>F(a,b,c,d).\tag{2}
$$

A. To $(a,b,c,d,e)$ be a solution, there must be
$$
F(a,b,c,d,e)-\dfrac{1}{abcde}\ge 2\tag{3},
$$
$$
F(a,b,c,d,e)> 2\tag{3'}.
$$
If $a\ge 7$, then
$$
F(a,b,c,d,e)\le F(a,a,a,a,a)\le\left(\dfrac{8}{7}\right)^5<2,
$$
contradiction with $(3')$.
So, if $(a,b,c,d,e)$ is a solution, then $$2\le a \le 6.\tag{4}$$

B. For each possible $a$ consider value 
$$
N_a = \lfloor F(a)\rfloor
$$
and focus on such $b$ only, that
$$F(a,b,b,b,b)> N_a+1.$$
For other $b$ one will have
$$
N_a\le F(a)<F(a,b)<F(a,b,c)<F(a,b,c,d)<F(a,b,c,d,e)-\dfrac{1}{abcde}\le F(a,b,b,b,b)-\dfrac{1}{abcde} < N_a+1,
$$
i.e. $F(a,b,c,d,e)-\dfrac{1}{abcde}$ is between two consecutive integers, so it cannot be integer.

C. For each possible pair $(a,b)$ consider value 
$$
N_b = \lfloor F(a,b)\rfloor
$$
and focus on such $c$ only, that
$$F(a,b,c,c,c)> N_b+1.$$
For other $c$ one will have
$$
N_b\le F(a,b)<F(a,b,c)<F(a,b,c,d)<F(a,b,c,d,e)-\dfrac{1}{abcde}\le F(a,b,c,c,c)-\dfrac{1}{abcde} < N_b+1.
$$

D. For each possible $(a,b,c)$ consider value 
$$
N_c = \lfloor F(a,b,c)\rfloor
$$
and focus on such $d$ only, that
$$F(a,b,c,d,d)> N_c+1.$$

E. For each possible $(a,b,c,d)$ consider value 
$$
N_d = \lfloor F(a,b,c,d)\rfloor
$$
and focus on such $e$ only, that
$$F(a,b,c,d,e)> N_d+1.$$

Applying described procedure, one will find four solutions:
$(a,b,c,d,e) = (2,4,16,256,65534)$, 
$(a,b,c,d,e) = (2,4,16,284,2506)$, 
$(a,b,c,d,e) = (4,4,4,42,5374)$, 
$(a,b,c,d,e) = (4,4,8,8,88)$; 
two of them have all distinct values $a,b,c,d,e$:
$$(a,b,c,d,e) = (2,4,16,256,65534),$$
$$(a,b,c,d,e) = (2,4,16,284,2506).$$
