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
 A: 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. 
A: Here's a line of attack that leaves you the task of checking only finitely many cases.
The idea is to check divisibility by powers of two. I use the notation $\nu(m)$ to denote
the exponent of the highest power of two that divides a positive integer $m$, so e.g. $\nu(80)=\nu(16\cdot5)=4$, because $2^4=16$.
You have probably seen (or can prove it as an exercise) that
$$
\nu(m!)=\sum_{i=1}^{\lceil \log_2m\rceil} \lfloor\frac m {2^i}\rfloor.\qquad(*)
$$
Also $\nu(mn)=\nu(m)+\nu(n)$, and consequently $\nu(m^n)=n\nu(m)$. Furthermore, if $\nu(a)\neq\nu(b)$ then $\nu(a+b)=\min\{\nu(a),\nu(b)\}$. This latter property is known as non-Archimedean triangle inequality. Let's call it NATI for short.
Here we see that $\nu(2014^{2015})=2015\cdot\nu(2014)=2015$ because $2014$ is even, but not divisibile by four.
Let us first consider the possibility that $a$ is odd. Then so is $a^{2014}$, so $2015\cdot b!$ must be odd. This is possible only, if $b!$ is odd, so we must have $b=0$
or $b=1$ for otherwise $b!$ is even. Check, whether there is matching $a$.
Let us then turn our attention to the possibility that $a$ is even. We first assume that $a$ is not divisible by four or, equivalently, that $\nu(a)=1$. In that case $\nu(a^{2014})=2014$. Because $\nu(2014^{2015})=2015$, NATI restricts the choice of $b$ to those numbers that have $2014=\nu(2015\cdot b!)=\nu(b!)$. Formula $(*)$ comes in handy. I give you the extra bit $\nu(2022!)=2014$. There are very few other possibilities for $b$, because $\nu(b!)$ is a (non-strictly) increasing function of $b$.
Finally, if $\nu(a)>1$, then $\nu(a^{2014})\ge2\cdot2014$, which is greater than $2015$. In this cases NATI forces $\nu(b!)=2015$. This again severely limits the possibilities for $b$.
A: Not the whole solution, but enough to get going.
We have that $\displaystyle a^{2014}+b!\equiv 0 \pmod{2014}$


*

*If $b\geq 53$, then $\displaystyle a^{2014}\equiv 0 \pmod{2014}$. 


Since $2014=2\times19\times53$, this implies $\displaystyle a\equiv 0 \pmod{2014}$
Hence $a=2014k$ for some positive $k$
Plugging this in yields $\displaystyle 2014^{2014}k^{2014}+2015  b!=2014^{2015}$
Hence $2014^{2014}(2014-k^{2014})=2015  b!$
The RHS being positive, so is the LHS, and thefore $2014\geq k^{2014}$, which forces $k=1$
Therefore $a=2014$. Furthermore, $2014^{2014}(2013)=2015  b!$ is clearly a contradiction.


*

*Thus, $b\leq 52$

A: Hint $ $  If $\,n\,$ is odd,  and $\ a^{n-1}\! + n b = (n-1)^{2j+1} $  then every prime $\,p\,$ dividing $\,n\,$ is $\,\equiv 1\pmod 4 $
since $\ n\!-\!1 = 2k,\,$ so $\, {\rm mod}\ p\!:\ (a^k)^2 \equiv\, -1,\ $ so by Euler or reciprocity, $ $ we infer $\ p\equiv 1\pmod 4$
