Problem 3. For natural numbers $x$ and $y$ and an odd integer $z$, it is known that $x! + y! = 8z + 2017$. 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: $(1; 4; -249), (4; 1; -249), (1; 5; -237), (5; 1; -237)$.
Solution. The expression $x!+y!$ must be an odd number, so $x=1$ or $y=1$.

Let $x=1$, then $1+y!=8z+2017$ or $y!=8z+2016$. Since $2016=8 \cdot 252$, $y!$ is divisible by 8, hence $y \geqslant 4$. If $y \geqslant 6$, then $y!$ is divisible by 16 and $2016=16 \cdot 126$. In this case, $z$ would be even, which contradicts the problem's condition. Therefore, $y=4$ or 5.