2,3,7 is the only triple $\geq 2$ that satisfies this property The property being that "taking the product of two of the numbers and adding one yields a number that is divisible by the third".
Clearly, this holds for 2,3,7 since


*

*$2\cdot 3 + 1 = 1\cdot 7$

*$2\cdot7 + 1 = 5\cdot 3$

*$3\cdot 7 + 1 = 11\cdot 2$


but why is this the only such triple?
This is from a (German) textbook on algebraic number theory, reviewing elementary number theory in chapter 1. Up to this point, only divisibility, the gcd and the euclidean algorithm have been introduced (no prime numbers yet). I took an elementary number theory course last semester, but I'm completely puzzled by this problem and don't know where to start. Trying to set up some equations and to manipulate them doesn't lead me anywhere.
My solution attempt
We can, without loss of generality, assume $2 \leq a \leq b \leq c$. From MXYMXY's answer below we have $abc|ab+bc+ca+1$ (which I'll denote (*)) and also that $a < 4$.
Let's then consider the two possibilities for $a$ in turn.


*

*$a =3$: We can show that $b$ and $c$ can't both be $\geq 4$, for if they were we would have $$\frac{3}{c}+1+\frac{3}{b}+\frac{1}{bc} \leq \frac{3}{4}+1+\frac{3}{4}+\frac{1}{16} < 3 \Leftrightarrow 3b + bc + 3c + 1 < 3bc,$$
but from (*) we have $3bc|3b+bc+3c+1$, which is a contradiction.
It follows that $b = 3$ and then we have $ab+1=10$, so $c|10$. Hence $c = 5$ or $c = 10$, but $a = 3$ divides neither $3\cdot5 + 1 = 16$ nor $3\cdot10 + 1 = 31$. Thus, this case is impossible.

*$a = 2$. We will show that $b$ and $c$ can't be both $\geq 5$. Otherwise we'd have $$\frac{2}{c}+1+\frac{2}{b}+\frac{1}{bc} \leq \frac{2}{5}+1+\frac{2}{5}+\frac{1}{25} < 2 \Leftrightarrow 2b + bc + 2c + 1 < 2bc,$$ but from (*) we have that $2bc|2b+bc+2c+1$, which is a contradiction. This implies that $2 \leq b < 5$. Let's consider the different cases for $b$:


*

*$b=4$: Then we have $ab+1=9$, so $c|9$, i.e. $c = 9$. But $a = 2$ does not divide $4\cdot 9 + 1 = 37$, so this is a contradiction.

*$b=2$: Then $ab+1=5$, so $c|5$, i.e. $c = 5$. But $a = 2$ does not divide $2\cdot 5 + 1 = 11$, so we also have a contradiction.



The only case left is where $a = 2, b = 3$. Then, we have $ab+1=7$, so $c|7$. Hence, $c=7$.
Hence, either $2,3,7$ is the only solution or there is none. That it is indeed a solution was verified above.
 A: HINT
Since $a|bc+1$, $b|ca+1$, $c|ab+1$, multiplying these together gives us that $$abc|ab+bc+ca+1$$ 
Note that $$a,b,c \ge 4 \Rightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{abc}<1 \Leftrightarrow abc>ab+bc+ca+1$$
This gives us that one of $a,b,c$ must be smaller than $4$. 
I think you can continue to divide cases from here. It requires slightly complicatd calculations, but not very. 
A: Equivalently $a$ divides $bc+1$ and cyclic ones. In particular, they are pairwise coprime. Then $abc$ divides $\prod_{\mathrm{cyc}}(ab+1)$, implying that $abc$ divides $1+\sum_{\mathrm{cyc}}ab$. Suppose wlog $a>b>c\ge 2$. If $c\ge 3$ then 
$$
3ab \le abc \le 1+\sum_{\mathrm{cyc}}ab\le 1+(ab-1)+(ab-1)+ab<3ab,
$$
which is impossible. Then $c$ has to be $2$. Then conditions are (i) $2\mid ab+1$ (i.e., $a$ and $b$ are odd), (ii) $a\mid 2b+1$ and (iii) $b\mid 2a+1$. Hence 
$$
b\le 2a+1 \le 2(2b+1)+1 \implies \frac{2a+1}{b} \in \{1,2,3,4,5\}.
$$
The ratio between odd numbers has to be odd, hence [$2a+1=3b$ or $5b$] and [$2b+1=a$ or $3a$]. Therefore we find the above unique solution $(2,3,7)$.
