
6. Determine all triples $(a, b, c)$ of positive integers such that $a \leq b \leq c$ and

$$
a+b+c+a b+b c+c a=a b c+1
$$


Solution: Putting $a-1=p, b-1=q$ and $c-1=r$, the equation may be written in the form

$$
p q r=2(p+q+r)+4
$$

where $p, q, r$ are integers such that $0 \leq p \leq q \leq r$. Observe that $p=0$ is not possible, for then $0=2(p+q)+4$ which is impossible in nonnegative integers. Thus we may write this in the form

$$
2\left(\frac{1}{p q}+\frac{1}{q r}+\frac{1}{r p}\right)+\frac{4}{p q r}=1
$$

If $p \geq 3$, then $q \geq 3$ and $r \geq 3$. Then left side is bounded by $6 / 9+4 / 27$ which is less than 1 . We conclude that $p=1$ or 2 .

Case 1. Suppose $p=1$. Then we have $q r=2(q+r)+6$ or $(q-2)(r-2)=10$. This gives $q-2=1, r-2=10$ or $q-2=2$ and $r-2=5$ (recall $q \leq r)$. This implies $(p, q, r)=(1,3,12)$, $(1,4,7)$.

Case 2. If $p=2$, the equation reduces to $2 q r=2(2+q+r)+4$ or $q r=q+r+4$. This reduces to $(q-1)(r-1)=5$. Hence $q-1=1$ and $r-1=5$ is the only solution. This gives $(p, q, r)=(2,2,6)$.

Reverting back to $a, b, c$, we get three triples: $(a, b, c)=(2,4,13),(2,5,8),(3,3,7)$.
