what is the value of $abc $ where $a > 1, b>1,c>1$ and $ (abc(abc-a-b-c)+ab+bc+ca-1) \equiv 0 \pmod{(abc)} $? Given that,
$a > 1, b>1,c>1,$
$a,b,c$ are distinct. 
and
$ (abc(abc-a-b-c)+ab+bc+ca-1) \equiv 0 \pmod{(abc)} $
Find the value of $abc$.

The only solutions are $(a,b,c)=(2,3,5)$ and its permutations, with $abc=30$. 
We only need to check triples $(a,b,c)$ with $abc\le ab+bc+ca-1$, because otherwise $abc$ cannot divide $ab+bc+ca-1$, and the congruence condition does not hold. This is a finite problem, easy to solve. For example, without loss of generality, we may assume that $a$ is the smallest number of $a,b,c$. For $a=2$ we see that we need $b=3$, and then we obtain $6c\le 5(c+1)$, which gives $c\le 5$, so that $c=5$ (because $c=4$ is impossible, and $a,b,c>1$ are distinct).
