Finding the integers

Question

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.

Answer 1

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 $(5)$.