USAJMO 2017 P4: triples $(a,b,c)$ such that $(a-2)(b-2)(c-2)+12$ is a prime number that divides the positive number $a^2+b^2+c^2+abc-2017$? 
Are there any triples $(a,b,c)$ of positive integers such that $(a-2)(b-2)(c-2)+12$ is a prime number that properly divides the positive number $a^2+b^2+c^2+abc-2017$?

My progress: Let $x=a-2$, $y=b-2$, $z=c-2$, then we get $xyz+12 \mid \left( x+y+z+4 \right)^2 - 45^2 \implies xyz+12 \mid (x+y+z-41)(x+y+z+49)$. So $xyz+12 \mid (x+y+z-41)$ or $xyz+12|(x+y+z+49)$ .
Also $xyz+12 $ can't divide both or else then $xyz+12 |90$ ( which doesn't work ).
WLOG $x\ge y\ge z$, then clearly $x\ge 9$ , since $11^2\cdot 3+ 11^3\le 2017$ ( not satisfying properly divide property).
Also I got $x,y,z$ odd or not divisible by $1 \mod 3$, or else $(a-2)(b-2)(c-2)+12$ is not prime.

If we continue from here: $\;\;\;xyz+12 \mid x+y+z-41\;\;\;$ or $\;\;\;xyz+12|x+y+z+49$.
We can assume $x\leq y\leq z$. We see that $x,y,z\notin \{0,2,3,4,6,8,9\}$ since $xyz+12$ is prime.

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*If $x>1$ then $x,y\geq 5$ and so $xyz+12\geq 25z+12$ so in

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*first case: $$25z+1\leq |x+y+z-41|\leq 3z+41\implies z\le 1$$ and thus no solution.


*second case: $$25z+1\leq x+y+z+49\leq 3z+49\implies z\leq 2$$ so no solution again.




*If $x=1$ then  $\;\;\;yz+12 \mid y+z-40\;\;\;$ or $\;\;\;yz+12|y+z+50$.

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*If $y\neq 1$ then $y\geq 5$. In first case we have $5z+12\leq 2z+40\implies z\leq 9$ so $z=7$ (and $y=5$ or $y=7$) or $z=5$ (and $y=5$). None works. In second case we have $5z+12\leq 2z+50\implies z\leq 12$. So $z\in\{11,7,5\}$ but none works.

*If $y=1$ then $\;\;\;z+12 \mid z-39\;\;\;$ then $z+12\mid 51 \implies z=5$ or $\;\;\;z+12|z+51$, then $z+12\mid 39$ so $z=1$.



*If $x=-1$ then we proccede similary like in a previous case...
