My $100$ post

Why do we need to mod17

For $x\leq 1$ we get solutions $(0,2015)$ and $(1,2016)$ Suppose that $x\geq 2$ and $y\geq 2017$ $(2^{2015}+1)^x-1=2^y-2^{2015}$ Hence $V_2((2^{2015}+1)^x-1)=2015$ By LTE $V_2(2^{2015})+V_2(2^{2015}+2)-1+V_2(x)=2015$ So $V_2(x)=0$ which means that $x$ is odd Reducing mod $9$ we get $6^x+5\equiv 2^y+1 (mod 9)$ As $x>1$ we have $6^x\equiv 0 (mod9)$ so $2^y\equiv 4 (mod 9)$ Hence $y\equiv 2 (mod 6)$ Reducing mod $7$ we get $5^x+4\equiv 2^y+1 (mod 7)$ Since $y\equiv 2 (mod 6)$ we have $2^y\equiv 4 (mod 7)$ So $5^x\equiv 1 (mod 7)$ which implies $6|x$ Contradiction with the fact that $x$ is odd so $(0,2015)$ and $(1,2016)$ are only solutions