All answers of $abcd=a+b+c+d-3$ in natural numbers Given $$abcd=a+b+c+d-3$$
what are all possible 4-tuples $(a, b, c, d)$?
I think that one answer lies within $(a,b,c,d)=(n,1,1,1)$ and all permutations of this answer? Is that right?
 A: That is correct.  Now you should prove that if any two of them are greater than $1$ the equation fails.  That shows you have found all the solutions.  Because of the symmetry you can permute the variables so that $a \le b \le c \le d$.  It makes it easier to talk about.  In your solution you would then only have $d \gt 1$ (if any of them are).  Now assume $c \gt 1$ and derive a contradiction.
A: There are obvious answers of the form $(n,1,1,1,)$ and its permutations as you have correctly pointed out.Let there be two numbers from $a,b,c,d$ greater than 1 , suppose a  and b  are greater than 1 and other two are equal to 1.We have now $ab=a+b-1$ , upon minor rearrangements you convert this to $(a-1)(b-1)=0$ , its follows from here that a or b or both have to be 0 , which is a contradiction because all of $a,b,c,d$ have to be natural numbers.
therefore all solutions are $(n,1,1,1),(1,n,1,1),(1,1,n,1),(1,1,1,n)$ , where n is a natural number.For the next two cases - Take the very first case a,b,c=2 d=1.(the lowest values greater than 1 which can be taken)This will not satisfy the equation since 8 is not equal to 4 subsequently for all higher numbers in this case the product part will be greater than sum part that is $abcd>a+b+c+d$ , therefore RHS will always be less than LHS and subtracting 3 will worsen the RHS . The next case where all of a,b,c and d are greater than 1 can be proven similarly (negative votes are welcome for this line of reasoning , I don't like this way of proving either  )
