Finding the integers 
Find all integers $a,b,c$ with $1<a<b<c$ such that $(a-1)(b-1)(c-1)$
  is a divisor of $abc-1$.

I cannot understand how to solve this. I would appreciate any help.

Assume $1<a<b<c$ and $(a-1)(b-1)(c-1)\mid abc-1$ and let $d=\frac{abc-1}{(a-1)(b-1)(c-1)}$.
Then from $$(a-1)(b-1)(c-1)<ab(c-1)=abc-ab<abc-1 $$
we see that $d\ge2$. 
Assume $a\ge 4$. Then $b\ge 5$, $c\ge 6$ and
$$abc-1=d(a-1)(b-1)(c-1)\ge d\cdot \frac34a\cdot\frac 45b\cdot \frac56c=\frac d2abc \ge abc,$$
contradiction! We conclude $$\tag1a\in\{2,3\}.$$
Note that $(a-1)(b-1)(c-1)$ also divides
$$(a-1)\cdot(abc-1)-a\cdot (a-1)(b-1)(c-1)=(a-1)\cdot(a(b+c)-a-1)$$
so that
$(b-1)(c-1)\mid  a(b+c)-a-1$
and
$$ b-1\mid \frac{a(b+c)-a-1}{c-1}=a+\frac{ab-1}{c-1}\le a+\frac{a(c-1)-1}{c-1}<2a$$
and in particular
$$\tag2c-1\mid ab-1$$
and $$\tag3a<b\le 2a.$$
One possibility compatible with $(2)$ is that $c=ab$. In that case  $(a-1)(b-1)(ab-1)$ divides $a^2b^2-1=(ab-1)(ab+1)$, i.e., $(a-1)(b-1)\mid ab+1$ and also $$(a-1)(b-1)\mid ab+1-(a-1)(b-1)=a+b.$$ For $a=2$ this means $b-1\mid b+2$ and so $b-1\mid (b+2)-(b-1)=3$, i.e., $b=4$. We obtain the solution
$$ \tag4a=2,\quad b=4,\quad c=8.$$
For $a=3$ instead we find $2(b-1)\mid b+3$, so $2(b-1)\mid 2(b+3)-2(b-1)=8$, $b-1\mid 4$, and ultimately $b=5$. We obtain a second solution
$$\tag5 a=3,\quad b=5,\quad c=15.$$
By $(1)$, those are all solutions with $c=ab$. Hence by $(2)$ we may assume from now on that $ab-1\ge 2(c-1)$. But we also have $\frac{ab-1}{c-1}\le \frac{a(c-1)-1}{c-1}<a\le 3$ so that in fact $$\tag6 ab-1=2(c-1).$$
In particular, $a,b$ must be odd so that from $(1)$ and $(3)$ we find $a=3$, $b=5$, and then from $(6)$ $c=8$. However, this does not provide a solution.
Hence the only solutions to the problem are given by $(4)$ and $(5)$.
