a + b + c + d = abcd
Thus
a + b + c = abcd - d
a + b + c = (abc - 1)d
d = (a + b + c)/(abc - 1)
Thus a + b + c must be divisible by abc - 1.
I am having a disagreeably difficult time meeting (even envisioning) this restriction unless
a + b + c = abc - 1
So
a + b + 1 = abc - c
a + b + 1 = (ab - 1)c
c = (a + b + 1)/(ab -1)
So a + b + 1 must be divisible by ab - 1.
I am going to again try
a + b + 1 = ab - 1
a + 2 = ab - b
a + 2 = (a - 1)b
b = (a + 2)/(a - 1)
So a + 2 is divisible by a - 1.
Now a + 2 is never equal to a - 1. So let
a + 2 = n(a - 1)
where n is also a natural number.
Then
a + 2 = na - n
a - na = - 2 - n
na - a = n + 2
(n - 1)a = n + 2
a = (n + 2)/(n - 1)
a must be a natural number, so let's try this for various values of n > 1:
n = 2 ==> a = 4
n = 3 ==> a = 5/2
n = 4 ==> a = 2
n = 5 ==> a = 7/4
n = 6 ==> a = 8/5
etc.
So it looks like a = 4 or a = 2.
Use a = 2.
Then
b = (a + 2)/(a - 1) = 4
c = (a + b + 1)/(ab -1) = 7/7 = 1
d = (a + b + c)/(abc - 1) = 7/7 = 1
So a = 2, b = 4, c = 1, d = 1.
(Check: 2 + 4 + 1 + 1 = 8 and 2*4*1*1 = 8.)
Using a = 4 we get b = 2, c = 1, d = 1 so this gives nothing new.
I can't find a way to get another set, though from the wording of the question more than one exists.
-Dan