I arrived at the diophantine problem for $3$ positive integers that should maintain all 3 divisibilities jointly:
$$ a | 1+a+b+c \\\
b | 1+a+b+c \\\
c | 1+a+b+c \tag 1 \\\
$$
I tried putting up a matrix expression introducing positive integer parameters $(i,j,k)$
$$\begin{array}{} & \\
\begin{bmatrix}
-i&1&1 \\
1&-j&1 \\
1&1&-k \\
\end{bmatrix} & \cdot \begin{bmatrix} a\\b\\c \end{bmatrix}
=\begin{bmatrix} -1\\-1\\-1 \end{bmatrix}
\end{array} \tag 2 $$
but the fiddling with the occuring formulae using $(i,j,k)$ now is again not yet conclusive.
I didn't find another convincing ansatz towards a formula.
By brute force ($2\le a \le b \le c \le 120$) I found the following solutions (avoiding symmetries) for $[1,a,b,c]$
[1, a, b, c]
--------------
[1, 2, 2, 5]
[1, 2, 3, 6]
[1, 2, 6, 9]
[1, 3, 4, 4]
[1, 3, 8, 12]
[1, 4, 5, 10]
[1, 6, 14, 21]
Searching using eq(2) with $1\le i \le j \le k \le 32$ I got
[i, j, k] ---> [1, a, b, c] rotated
-------------------------------
[1, 2, 6] ---> [1,21,14, 6] R
[1, 2, 7] ---> [1,12, 8, 3] R
[1, 2, 8] ---> [1, 9, 6, 2] R
[1, 2,11] ---> [1, 6, 4, 1]
[1, 3, 4] ---> [1,10, 5, 4] R
[1, 3, 5] ---> [1, 6, 3, 2] R
[1, 3, 7] ---> [1, 4, 2, 1]
[1, 4, 4] ---> [1, 5, 2, 2] R
[1, 5, 5] ---> [1, 3, 1, 1]
[2, 2, 3] ---> [1, 4, 4, 3] R
[2, 2, 5] ---> [1, 2, 2, 1]
[3, 3, 3] ---> [1, 1, 1, 1]
which are simply rotated versions or trivials (having $a or b or c=1$) which are excluded in the above list.
I guess these are all possible solutions, but don't find the argument…
Q1: How can I approach that problem algebraically?
Q2: is the number of solutions finite or infinite?
Q3.1: if the set of solutions is infinite is there a parametrization?
Q3.2: if the set of solutions is finite, what is that set?
update
In generalizing the problem towards [h,a,b,c] with $h \in \mathbb N^+$ I seem to get the full list of $14$ solutions (avoiding solutions with $\gcd()>1$ and rotations) which are
[h, a, b, c] some interpretations
--------------------------------------------
[1, 1, 1, 1]
[1, 1, 1, 3]
[1, 1, 2, 2]
[1, 1, 2, 4]
[1, 1, 4, 6]
[1, 2, 2, 5] 1+2+2=5: 5=5/1
[1, 2, 3, 6] 1+2=3: 3=1*3 6=2*3
[1, 2, 6, 9] 1+2=3: 6=2*3 9=3*3
[1, 3, 4, 4] 1+3=4: 4=1*4 4=1*4
[1, 3, 8, 12] 1+3=4: 8=2*4 12=3*4
[1, 4, 5, 10] 1+4=5: 5=1*5 10=2*5
[1, 6, 14, 21] 1+6=7: 14=2*7 21=3*7
[2, 3, 3, 4] 2+3+3=8: 4=8/2
[2, 3, 10, 15] 2+3=5: 10=2*5 15=3*5
A bit context: this is a detail-problem in an earlier question where I explore the general conditions in the Lehmer's totient problem. In the earlier question I've considered a three-variable $(R,S,T)$ diophantine system, and looked at solutions of the form $(R,S,T)=(R^1,R^a,R^b)$ Here I generalize it to $4$ variables and solutions of $(Q,R,S,T)=(Q^1,Q^a,Q^b,Q^c)$ resp. $(Q,R,S,T)=(Q^h,Q^a,Q^b,Q^c)$ ($\gcd(h,a,b,c)=1$) and determine the solutions in terms of $(1,a,b,c)$ resp. $(h,a,b,c)$. I'll later generalize to more variables but first I want to get some grip about the general limitations and whereabouts – at best in a form which support later generalizations…
Best Answer
Because of symmetry, you can assume that $a\leq b\leq c$, then we have $$ck= 1+a+b+c \implies k\leq 4$$
If $\boxed{k=2}$ we have $2c=1+a+b+c$ so $c= 1+a+b$. Then $b\mid 2+2a+2b$ so $b\mid 2+2a$ so $mb = 2a+2\leq 4b\implies m\leq 4$ and now is easy to check all 4 subscases...
If $\boxed{k=3}$ we have $3c=1+a+b+c$ so $2c=1+a+b$.
If $b\leq c-1$ then $2c\leq 1+2c-2 = 2c-1$ a contradiction.
If $b= c$ then $c=1+a$ and now we have $a\mid 3+3a \implies a\mid 3$ so $a=1$ and $b=c=2$ or $a=3$ and $b=c=4$.