Integer solutions to $a^{2014} +2015\cdot b! = 2014^{2015}$ 
How many  solutions are there for $a^{2014} +2015\cdot b! = 2014^{2015}$, with $a,b$ positive integers?

This is another contest problem that I got from my friend.
Can anybody  help me  find the answer? Or  give me a hint to solve this problem?
Thanks

Taking this equation mod $2015$ yields $a^{2014} \equiv -1 \pmod{2015}$. 
Since $2015 = 5 \cdot 13 \cdot 31$, we get the following: 
$a^{2014} \equiv -1 \pmod{5}$
$a^{2014} \equiv -1 \pmod{13}$
$a^{2014} \equiv -1 \pmod{31}$
By Fermat's Little Theorem, $a^{31} \equiv a \pmod{31}$. Hence, $a^4 \equiv a^{2014} \equiv -1 \pmod{31}$. 
We can check that $-1$ is not a quadratic residue $\pmod{31}$. Thus, there is no residue $a^2$ such that $(a^2)^2 = a^4 \equiv -1 \pmod{31}$. Therefore, there are no solutions to the original equation. 
