Mongolian Mathematical Olympiad

Answer: $(a,b,c)=(5,8,14)$. We prove that this is the only solution. Consider the equation modulo $16$. We have $c^4+2024\equiv 8,9(\mathrm{mod}16)$ on the right-hand side, since $x^4\equiv 0,1(\mathrm{mod}16)$. Now we consider the left-hand side. First, we have $b\ge 7$, otherwise $2\cdot 6!=1440<2024$. Hence $b!\equiv 0(\mathrm{mod}16)$. Now, $(1!,2!,3!,4!,5!,6!)(\mathrm{mod}16)=(1,2,6,8,8,0)(\mathrm{mod}16)$, therefore $a=4,5$ and $c=2d$ even. Note that $7!/16=315$. For $a=4$, we have $b!/16=d^4+125\equiv 5,6(\mathrm{mod}8)$ but $7!/16\equiv 3(\mathrm{mod}8)$ and $b!/16\equiv 0(\mathrm{mod}8)$ for $b\ge 8$. This means $a=4$ is impossible. For $a=5$, we have $b!/16=d^4+119\equiv 7,8(\mathrm{mod}16)$, but $7!/16\equiv 11(\mathrm{mod}16)$ and $b!/16\equiv 0(\mathrm{mod}16)$ for $b\ge 10$. Thus the only possibilities are $b=8,9$. It is easy to check that $b=8$ is a solution with $c=14$, and $b=9$ is not a solution.