https://www.artofproblemsolving.com/community/c3046h1184227_the_system_is_an_arithmetic_progression

Any ideas?

Let $a>b>c$ $\frac{st}{r}$ is integer , so $\frac{s^2t^2}{r^2}$ is integer too. $\frac{s^2t^2}{r^2}=\frac{(ac+1)(bc+1)}{ab+1}=\frac{abc^2+c(a+b)+1}{ab+1} =\frac{(ab+1)(c^2+1)+(ac+bc-c^2-ab)}{ab+1} = c^2+1-\frac{(a-c)(b-c)}{ab+1}$ $(a-c)(b-c)>0$ and $ab+1|(a-c)(b-c)$ so $(a-c)(b-c)\geq ab+1$ $ab-ac-bc+c^2 \geq ab+1$ $c^2\geq ac+bc+1 > 2c^2+1$ - contradiction. So $\frac{st}{r}$ is not integer

Here's a similar solution to the above. Assume otherwise. Note that if $\dfrac{rt}{s}$ is an integer, then $\dfrac{r^2s^2}{t^2}$ must be too. Therefore, $\dfrac{(ab+1)(bc+1)}{ac+1}$ is an integer, and so $(ac+1) \mid (ab^2c+ab+bc+1)$. Note that $ab^2c \equiv -b^2 \pmod {ac+1},$ and so $(ac+1) \mid (ab+bc+1-b^2)$. Therefore, $ab+bc+1 -b^2 \geq ac+1,$ hence $ab+bc-ac \geq b^2$. Summing the cyclic inequalities we get $ab+bc+ca=(ab+bc-ca)+(bc+ca-ab)+(ab+ac-bc) \geq a^2+b^2+c^2,$ and so $a=b=c$, which contradicts the assumption.