USAJMO 2017 Problem 4

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.

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

  • 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 $\;\;\;yz+12|y+z+50$.

  • 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...