The only solutions are $(a,b,c)=(2,4,8)$ and $(a,b,c)=(3,5,15)$ Let $a,b,c$ such that $1 <a <b <c$ and $abc-1$ divisible by $(a-1)(b-1)(c-1)$: $1<\frac{abc-1}{(a-1)(b-1)(c-1)}<\frac{a}{a-1}\frac{b}{b-1}\frac{c}{c-1}\leq2.\frac{3}{2}.\frac{4}{3}=4$ Thus $\frac{abc-1}{(a-1)(b-1)(c-1)}=2$ or $3$ We have $a=2$ or $3$ because if $a>3$ then $\frac{a}{a-1}\frac{b}{b-1}\frac{c}{c-1}\leq\frac{4}{3}.\frac{5}{4}.\frac{6}{5}=2$. It's easy to study the two cases for $a$... _____________ blang (sorry for my french english)

Actually, ehsan2004 asked us "is it easy or not", he didn't asked us to solve the problem So I can answer to his question: IT IS EASY

Let $M=\frac{abc-1}{(b-1)(c-1)}$ is natural, which is divisible by $a-1$. Now we will write the intervals of $M$. 1) $M>\frac{abc-ac}{(b-1)(c-1)}=\frac{ac}{c-1}>a$. So we have $M\geq a+1$. 2) $M<\frac{abc}{(b-1)(c-1)}=(a+\frac{a}{b-1})\frac{c}{c-1}\leq (a+1)(\frac{c}{c-1} =a+1+\frac{a+1}{c-1}\leq a+2$. Which implies to $M=a+1$. As we know $\frac{M}{a-1}=1+\frac{2}{a-1}$ is natural thus a=2 or a=3. Two more cases: 1) If a=2 so from $M=a+1$ ew have $c=3+\frac{5}{b-3}$, from here we get b=4,c=8. 2) If a=3 analogously $c=4+\frac{11}{b-4}$ and from this we get b=5,c=15. davron

A bit of a different approach that's lengthier:

I think it's a new solution for this problem.

The most difficult process: ${(a-1)(b-1)(c-1) \mid abc-1-(a-1)(b-1)(c-1)=\sum{ab}-\sum{a} \Rightarrow c \le \frac{1-a-b+2ab}{ab-2(a+b)+1}}$. We may now see that $a \ge 7 \Leftrightarrow 2ab-7(a+b)+3 \ge 14(a+b)-98-7(a+b)+3=7(a+b)-95 \ge 0 \Leftrightarrow c=\frac{1-a-b+2ab}{ab-2(a+b)+1} \le 4$.(The second line follows from the fact that $(a-7)(b-7) \ge 0$ Contradiction.So we are left to check the five cases $a=2,3,4,5,6$,which is long but not too boring.

Myth wrote: Actually, ehsan2004 asked us "is it easy or not", he didn't asked us to solve the problem So I can answer to his question: IT IS EASY o pidr kimga tawavossan abjag'ini chiqarib qoyaman sangamas bunaqa gaplani qiliw

it is very easy problem =a-1 y=b-1 z=c-1

mortezamoradi wrote: I think it's a new solution for this problem.

Correct

Nice.......

Solution from OTIS office hours: The answers are $(2,4,8)$ and $(3,5,15)$ which are seen to work. Claim: The numbers $(a,b,c)$ have the same parity. Proof. Otherwise $(a-1)(b-1)(c-1)$ is even but $abc-1$ is odd. $\blacksquare$ Claim: We have $a < 4$. Proof. Since \[ 2 < \frac{abc}{(a-1)(b-1)(c-1)} < \frac{a}{a-1} \cdot \frac{a+2}{a+1} \cdot \frac{a+4}{a+3} \]this implies the result. $\blacksquare$ At this point let $x=a-1$, $y=b-1$, $z=c-1$; then the condition rewrites as \[ xyz \mid xy+yz+zx+x+y+z \]and we know $(x,y,z)$ are same parity. We consider two cases. If $x=1$ then the equation reads \[ yz \mid 2(y+z)+1. \]This fails if $yz > 2(y+z)+1 \iff (y-2)(z-2) > 5$, so check $(y,z) \in \left\{ (3,5), (3,7) \right\}$. If $x=2$ then let $y=2r$ and $z=2s$; we get the equation \[ 4rs \mid 2rs + 3r + 3s + 1. \] If $4rs = 2rs+3r+3s+1$ then $(2r-3)(2s-3) = 11$, which gives $r=2$ and $s=7$, which works. Otherwise we have $8rs > 2rs+3r+3s+1$, so the equation fails.

Using a function and bounding that makes this problem a casework I think I have done the same problem in some imo i don't remember the year

The only solutions are $(2, 4, 8)$ and $(3, 5, 15)$ which both work. Claim: $a<4.$ Proof: For $a \ge 4,$ we have $$1 < \frac{abc-1}{(a-1)(b-1)(c-1)} < \frac{a}{a-1} \cdot \frac{b}{b-1}\cdot \frac{c}{c-1} \le \frac{4}{3} \cdot \frac{5}{4} \cdot \frac{6}{5}=2.$$This is a contraction because there are no integers between $1$ and $2.$ $\Box$ Now, we consider $2$ cases. Case 1: $a=2.$ We want $$\frac{2bc-1}{(b-1)(c-1)} \in \mathbb{N}, 2<b<c.$$We have $$1 < \frac{2bc-1}{(b-1)(c-1)}<\frac{2bc}{(b-1)(c-1)} \le 2 \cdot \frac{3}{2} \cdot \frac{4}{3}=4.$$So, we must have $\frac{2bc-1}{(b-1)(c-1)}=3,$ since its always odd. If $b \ge 6,$ we have $1< \frac{2bc-1}{(b-1)(c-1)} < \frac{2bc}{(b-1)(c-1)} \le 2 \cdot \frac{6}{5} \cdot \frac{7}{6}=\frac{14}{5} <3.$ So, $b<6 \implies b \in \{3, 4, 5\}.$ Subcase 1: $b=3.$ If $b=3,$ we have $$\frac{6c-1}{2c-2}=3 \implies 6c-1=6c-6 \implies \text{ contradiction.}$$ Subcase 2: $b=4.$ If $b=4,$ we have $$\frac{8c-1}{3c-3}=3 \implies 8c-1=9c-9 \implies c=8 \implies (a, b, c)=(2, 4, 8).$$ Subcase 3: $b=5.$ If $b=5,$ we have $$\frac{10c-1}{4c-4}=3 \implies 8c-1=12c-12 \implies c=\frac{11}{2} \implies \text{ contradiction.}$$Now if you follow the same steps for the other case, you will end up with the solution $(3,5,15),$ and these are the only $2$ solutions. $\blacksquare$

Answers are $\boxed{(2,4,8)}$ and $\boxed{(3,5,15)}$. These work. Lemma: $a<4$. Proof: Suppose $a\ge 4$. Then \[1<\frac{abc-1}{(a-1)(b-1)(c-1)}< \frac{abc}{(a-1)(b-1)(c-1)}\le \frac{4}{3}\cdot \frac{5}{4}\cdot \frac{6}{5}\le 2,\]a contradiction a no integer is between $1$ and $2$, exclusive. $\blacksquare$ Case 1: $a=3$. Then $2(b-1)(c-1)=2bc-2b-2c+2\mid 3bc-1$. Note that \[6bc-6b-6c+6-(3bc-1)=3bc-6b-6c+7=3(b-2)(c-2)-5>0,\]so $4bc-4b-4c+4=3bc-1$. This implies \[bc-4b-4c+5=0\implies (b-4)(c-4)=11\implies b=5, c=15\] Case 2: $a=2$. Subcase 1: $b>4$. Then $(b-1)(c-1)=bc-b-c+1\mid 2bc-1$. Note that \[3bc-3b-3c+3-(2bc-1)=bc-3b-3c+4=(b-3)(c-3)-5>0,\]so $2(b-1)(c-1)=2bc-1$, a contradiction. Subcase 2: $b=3$. Then $2c-2\mid 6c-1$. So \[2c-2=\gcd(2c-2,6c-1)=\gcd(2c-2,6c-1-(4c-4))=\gcd(2c-2,2c+3)\mid 5,\]contradiction. Subcase 3: $b=4$. Then $3c-3\mid 8c-1$. So \[3c-3=\gcd(3c-3,8c-1)=\gcd(3c-3,8c-1-(6c-6))=\gcd(3c-3,2c+5)=\gcd(2c+5,c-8)=\gcd(c-8,c+13)\mid 21\] So $c-1\mid 7\implies c=8$. We have exhausted all cases, so we are done.

Note first that: $$1>\frac{abc-1}{(a-1)(b-1)(c-1)}<\frac a{a-1}\cdot\frac b{b-1}\cdot\frac c{c-1}\le2\cdot\frac32\cdot\frac43=4$$since $a\ge2$, $b\ge3$, and $c\ge4$, from which it follows that either $\frac{abc-1}{(a-1)(b-1)(c-1)}=2$ or $\frac{abc-1}{(a-1)(b-1)(c-1)}=3$. Suppose that $a>3$. Then we obtain the stronger bound: $$\frac{abc-1}{(a-1)(b-1)(c-1)}<\frac a{a-1}\cdot\frac b{b-1}\cdot\frac c{c-1}\le\frac43\cdot\frac54\cdot\frac65=2,$$a contradiction. So either $a=2$ or $a=3$. Case 1: $a=2$ Then $\frac{2bc-1}{(b-1)(c-1)}$ is an integer, which implies that $b$ and $c$ are both even since the numerator is odd. So $b\ge4$ and $c\ge6$. Also, since $\frac{2bc-1}{(b-1)(c-1)}$ is odd, it must be $3$. By SFFT, $(a,b,c)=\boxed{(2,4,8)}$. Case 2: $a=3$ Then $\frac{3bc-1}{2(b-1)(c-1)}$ is an integer. Since $$\frac{3bc-1}{2(b-1)(c-1)}<\frac32\cdot\frac b{b-1}\cdot\frac c{c-1}\le\frac32\cdot\frac43\cdot\frac54<3$$the integer must be $2$. Finally SFFT produces the only other solution $(a,b,c)=\boxed{(3,5,15)}$.

Let \begin{align*} x&=a-1\\ y&=b-1\\ z&=c-1\\ 0<x<y<z \end{align*} $$xyz \vert xyz + xy + xz + yz + x + y + z$$, So we find all $x,y,z$ with $xyz \vert xy + xz + yz + x + y + z$ Here an important step is to note the difference in degree and the contrary inequality, $$xyz \leq xy + xz + yz + x + y + z$$Which can be used for bounding. Since $x,y,z$ are distinct positive integers, $$xy + xz + yz + x + y + z \leq \dfrac{17}{6} xyz$$With equality when $x=1,y=2,z=3$. Thus it suffices to consider $$xyz = xy + xz + yz + x + y + z$$And $$2xyz = xy + xz + yz + x + y + z$$ Consider $xyz = xy + xz + yz + x + y + z$ If $x \geq 6$, $xyz \geq 6yz > xy + xz + yz + x + y + z$, no soln. If $x=5$, we want to solve $4yz=6y+6z+5$. However $4yz \geq 24z > 6y+6z+5$, so no soln. If $x=4$, we want $3yz=5y+5z+4$. However $3yz \geq 15z > 5y+5z+4$, so no soln. If $x=3$, we want $2yz=4y+4z+3$. However $2yz \geq 8z \geq 4y+4z+4 > 4y+4z+3$, no soln. If $x=2$, we want $yz=3y+3z+2$. If $y \geq 6$, $yz \geq 6z \geq 3y+3z+3 > 3y+3z+2$, no soln. If $y=5$, $5z=17+3z$, $z \notin \mathbb{Z}$, no soln. If $y=4$, $4z=14+3z$, giving $z=14$ and the soln set (2,4,14). If $y \leq 3$, $yz \leq 3z < 3z + 3y + 2$, no soln. If $x=1$, $0=2y+2z+2$, no soln. Now consider $2xyz=xy+xz+yz+x+y+z$. If $x \geq 3$, $2xyz \geq 6yz > xy+xz+yz+x+y+z$, no soln. If $x=2$, we want $3yz=3y+3z+2$ However $3yz \geq 9z > 3y+3z+2$, no soln. If $x=1$, we want $yz=2y+2z+1$. If $y \geq 4$, $yz \geq 4z \geq 2z+2y+2 > 2y+2z+1$, no soln. If $y=3$, $3z=2z+7$ gives $z=7$, and the soln set $(1,3,7)$. If $y \leq 2$, $yz \leq 2z < 2y+2z+1$, no soln. Thus the only solns to $(x,y,z)$ is $$(x,y,z) = (2,4,14), (1,3,7)$$, Corresponding to $$(a,b,c) = (3,5,15), (2,4,8)$$. $\square$

Expand $(a-1)(b-1)(c-1)$ into $abc-ab-bc-ac+a+b+c-1$ Then, $\frac{abc-1}{abc-ab-bc-ac+a+b+c-1}$ simplifies to $1+\frac{ab+bc+ac-a-b-c}{abc-ab-bc-ac+a+b+c-1}$ $\frac{ab+bc+ac-a-b-c}{abc-ab-bc-ac+a+b+c-1}$ then simplifies to $\frac{a(b-1)+b(c-1)+c(a-1)}{(a-1)(b-1)(c-1)}$ Even more simplifying gives $\frac{a}{(a-1)(c-1)}+\frac{b}{(a-1)(b-1)}+\frac{c}{(b-1)(c-1)}$, which is less than $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ which is less than $3$ so $\frac{abc-1}{(a-1)(b-1)(c-1)}$ is less than $4$ but also greater than $1$. Therefore, $\frac{abc-1}{(a-1)(b-1)(c-1)}$ is equal to $2$ or $3$. Solve we get that $a=2$ or $a=3$ since $a>3 \implies \frac{abc-1}{(a-1)(b-1)(c-1)}<2$ which makes $\frac{abc-1}{(a-1)(b-1)(c-1)}$ neither equal to $2$ nor equal to $3$. Plugging in $a=2$, we get $b=4$ and $c=8$ by solving $\frac{2bc-1}{(b-1)(c-1)}=3$ and no solutions for $\frac{2bc-1}{(b-1)(c-1)}=2$ Plugging in $a=3$, we get $b=5$ and $c=15$ by solving $\frac{3bc-1}{2(b-1)(c-1)}=3$ and no solutions for $\frac{3bc-1}{2(b-1)(c-1)}=2$ Therefore, the only solutions for $(a,b,c)$ are $(2,4,8)$ and $(3,5,15)$

I resolved this one. The answer is $\boxed{(2,4,8)}$ and $\boxed{(3,5,15)}$, which work. Let $x = a-1, y = b-1, z = c-1$. We have $xyz \mid xy + yz + zx + x + y + z$. First note that $xy > x + y + 1$ for any integers $x \ge 3, y \ge 3$ because it's true for $(3,3)$ and stays true when you add $x$ or $y$ by $1$. Claim: We have $xyz > xy + yz + zx + x + y + z $ whenever $x \ge 3$. Proof: This is because it's true for $x = 3, y = 4, z = 5$, and adding $1$ to any variable keeps it true since it changes the LHS by $xy$ (for example) and the RHS by $x + y + 1$, and $xy > x + y + 1$ for $x, y \ge 3$. $\square$ This means that either $x = 1$ or $x = 2$. Case 1: $x=1$ Then $yz\mid 2(y+z) + 1$. We can do a similar trick as above for proving that $yz > 2(y+z) + 1$ for $y\ge 4$ since it's true for $(4,5)$ and when you increase either variable by $1$ it stays true. Therefore, either $y = 2$ or $y = 3$. If $y =2$, then $2z \mid 2z + 5$, bad. If $y = 3$, then $3z\mid 2z + 7$, so we must have $z = 7$, which works. Hence we have $(x,y,z) = (1,3,7)$, which corresponds to $(a,b,c) = (2,4,8)$. $\square$ Case 2: $x = 2$. Then $yz \mid 3(y+z) + 2$. We see that $yz > 3(y+z) + 2$ for $(y,z) = (6,7)$ and for $y,z$ that large, increasing either variable by $1$ results in an increase more than in the RHS (3), so it's true for any $y \ge 6$. Hence we must have $y\in \{3,4,5\}$. If $y = 3$, then $3z\mid 3z + 11$, bad. If $y = 4$, then $4z\mid 3z + 14$, so $z\mid 14$ and $z\equiv 2\pmod 4$, so $z = 14$. If $y = 5$, then $5z\mid 3z + 17\implies z = 17$, which doesn't work. Therefore, $(x,y,z) = (2,4,14)$, so $(a,b,c) = (3,5,15)$.

I claim the only solutions are $\boxed{(a, b, c) \in \{(2, 4, 8), (3, 5, 15)\}}$. As $a \geq 2$, we must have $b \geq 3$ and consequently $c \geq 4$. Observe that \[1 < \frac{abc - 1}{(a - 1)(b - 1)(c - 1)} < \frac{abc}{(a - 1)(b - 1)(c - 1)} = \left(1 + \frac{1}{a - 1} \right) \left(1 + \frac{1}{b - 1} \right) \left(1 + \frac{1}{c - 1} \right) \leq 2 \cdot \frac{3}{2} \cdot \frac{4}{3} = 4.\]Now make the astute observation that if $a \geq 4$, one obtains a contradiction: \[\left(1 + \frac{1}{a - 1}\right) \left(1 + \frac{1}{b - 1} \right) \left(1 + \frac{1}{c - 1} \right) \leq \frac{4}{3} \cdot \frac{5}{4} \cdot \frac{6}{5} = 2,\]which is a contradiction as there cannot be an integer between $1$ and $2$. Hence $2 \leq a \leq 3$. We now proceed to casework with all of our hitherto obtained information in mind: Case 1: $\left(\frac{abc - 1}{(a - 1)(b - 1)(c - 1)} = 2 \right).$ Then $abc - 1 = 2(a - 1)(b - 1)(c - 1)$. Subcase 1.1: $(a = 2).$ \begin{align*} 2bc - 1 &= 2(b - 1)(c - 1) \\ \implies 2bc - 1 &= 2bc - 2b - 2c + 2 \\ \implies 2(b + c) &= 3, \end{align*}which is impossible as the left-hand side is even, whereas the right-hand side is odd. $\bigstar$ Subcase 1.2: $(a = 3).$ \begin{align*} 3bc - 1 &= 2(2)(b - 1)(c - 1) \\ \implies bc - 4b - 4c + 5 &= 0 \\ \implies (c - 4)(b - 4) &= 11 \implies \boxed{c = 15, b = 5}, \end{align*}which yields a valid solution. $\bigstar$ Case 2: $\left(\frac{abc - 1}{(a - 1)(b - 1)(c - 1)} = 3 \right).$ Subcase 2.1: $(a = 2).$ \begin{align*} 2bc &= 3(1)(b - 1)(c - 1) \\ \implies bc - 3b - 3c + 4 &= 0 \\ \implies (c - 3)(b - 3) &= 5 \\ \implies \boxed{c = 8, b = 4}, \end{align*}which yields a valid solution. $\bigstar$ Subcase 2.2: $(a = 3).$ \begin{align*} 3bc - 1 &= 3(2)(b - 1)(c - 1) \\ \implies 3bc - 6b - 6c + 7 &= 0, \end{align*}which is impossible as the left-hand side is $1$ modulo $3$, and the right-hand side is $0$ modulo $3$. $\bigstar$. Therefore the only solutions are that of $\boxed{(a, b, c) \in \{(2, 4, 8), (3, 5, 15)\}}$, as desired. $\blacksquare$