2015 Happy New Year Special

For any pair of numbers a and b we know that $a^2-b^2=(a+b)(a-b)$ , so $2015^2-2014^2=(2015+2014)(2015-2014)=(2015+2014)(1)=2015+2014$ . Repeat for 2013 and 2012, 2011 and 2010, etc, and you find that the sum of the sequence is the sum of every number from 1 to 2015. Apply Gauss' formula $\frac{n(n+1)}{2}$ and you find that the sum of the first 2015 numbers is $\frac{2015\times 2014}{2}=2015\times 1007$ , which is obviously $0\pmod{2015}$ .

First,notice that $2015^2\equiv 0\pmod{2015}\\-(2014)^2\equiv -1\pmod{2015},1^2\equiv 1\pmod{2015}\\ 2013^2\equiv 2^2\equiv 4\pmod{2015},-(2^2)\equiv -4\pmod{2015}\\ -(2012^2)\equiv -3^2\equiv-9\pmod{2015},3^2\equiv 9\pmod{2015}\\.\\.\\.\\.\\ -(1008^2)\equiv -(1007)^2\pmod{2015},1007^2\equiv1007^2\pmod{2015}$ .When we add all the above expressions we have $2015^2-2014^2+2013^2........+3^2-2^2+1^2\equiv 0\pmod{2015}$ .