Now you may think that I am annoying, but if I am not asking this question, then it seems not so complete and I can't grasp the whole idea... refer to this question: Positive rationals satisfying $a+b+c+d=abcd$?
My question is for integer $a,b,c,d,e$, how do you find all solutions that satisfy
$$b+c+d+e=a(bcde-1)\;?$$
Sort a,b,c,d,e by decreasing absolute value : $|a| \ge |b| \ge |c| \ge |d| \ge |e|$.
If $e=0$ then the equation simplifies to $a+b+c+d = 0$.
If no integer is $0$, then $|a+b+c+d+e| \le 5|a|$, so $|bcde| \le 5$. So there is a finite number of possibilities for $(b,c,d,e)$.
Then, since $b+c+d+e = a(bcde-1)$, either $bcde=1$ and $b+c+d+e = 0$, which is only possible if $b,c,d,e = +1,+1,-1,-1$, and then $a$ can have any value ; either $bcde \neq 1$ and then you have to check if you get an integer when you solve for $a$.
So you get two infinite families of solutions, the family of solutions with a $0$, the family of solutions with two $1$ and two $-1$, and then you have a finite number of extra solutions that you can enumerate with a program.