Should It wait till next year?

Obviously,we need to look at the expression mod $20$ .Since $20=5\cdot 4$ , we look at the expression modulo $4$ and $5$ and see that $x$ must be divisible by both $2$ and $5$ ,and hence by $10$ .This implies that $x=10y$ for some positive integer $y$ .Next,we substitute it in the original expression,take mod $20$ and prove that all numbers divisible by $10$ and smaller then $2016$ satisfy the given property.Since the numbers divisible by $10$ form an arithmetic progression,it is easy to find the sum.

From the first look at the problem we can see that the best way to solve this problem is to use modular arithmetic,more specifically (mod20) and that'x' needs to be an integer because if 'x' is not an integer then the remainder will not be an integer.From the question we have that $x^{2015}-2015^{x}\equiv15(mod20).$ Checking a few cases we get that any power of 2015 will leave remainder either 15 or 5 when divided by 20.Now,let us do this case by case.Case 1: $2015^{x}\equiv15(mod20).$ Thus, $x^{2015}\equiv30\equiv10(mod20).$ But after a lot of hit and trial we get that there are no integer solutions to it.Does somebody have an easier way to do this part??Case 2: $2015^{x}\equiv5(mod20).$ Thus, $x^{2015}\equiv20\equiv0(mod20)$ This can happen when $a)x\equiv0(mod20)\ or\ b)x\equiv10(mod20).$ For a) we have numbers from $20*1-----20*100$ and for b) we have numbers from $20*0+10-----20*100+10.$ Now we can easily calculate the sums as all the numbers in both cases form A.Ps.