Problem 3. For natural numbers $x$ and $y$ and an odd integer $z$, it is known that $x! + y! = 16z + 2017$. Find all possible such triples of numbers $(x, y, z)$. (Recall that $1! = 1, 2! = 1 \cdot 2, n! = 1 \cdot 2 \cdot \ldots \cdot n$)
Answer: $(1 ; 6 ;-81),(6 ; 1 ;-81),(1 ; 7 ; 189),(7 ; 1 ; 189)$.

Solution. The expression $x!+y!$ must be an odd number, so $x=1$ or $y=1$.

Let $x=1$, then $1+y!=16z+2017$ or $y!=16z+2016$. Since $2016=16 \cdot 126$, $y!$ is divisible by 16, hence $y \geqslant 6$. If $y \geqslant 8$, then $y!$ is divisible by 32 and $2016=32 \cdot 63$. In this case, $z$ would be even, which contradicts the problem's condition. Therefore, $y=6$ or 7.