Denote \(\left\{{}\begin{matrix}A=\dfrac{2017^x-2016^{y+1}}{2015}\\B=2017^x-2016^{y+1}\\C=2015\end{matrix}\right.\)

For A is a square number then \(B⋮C\) 

<=> \(2017^x-2016^{y+1}⋮2015\)

<=> \(2017^x-2^x+2^x-2016^{y+1}+1-1⋮2015\)

<=> \(2^x-1⋮2015\)

<=> \(2^x\equiv1\left(mod2015\right)\)                                                                (*)

     +) With x = 0 then (*) is true 

     +) With x = 1 then (*) is wrong

     +) With x = 2 then (*) is wrong

     +) With \(x\ge3\) => \(\left[{}\begin{matrix}x=3k\\x=3k+1\\x=3k+2\end{matrix}\right.\left(k>0\right)\)

           -) x = 3k => (*) <=> \(2^{3k}\equiv1\left(mod2015\right)\)  <=>  \(8^k\equiv1\left(mod2015\right)\)  (wrong , because k > 0)

           Similar , we can show that with x = 3k + 1 , x = 3k + 2 then (*) is still wrong 

So x = 0 is satisfy 

Change x = 0 into (*) 

=> \(1-2016^{y+1}⋮2015\) 

For \(1-2016^{y+1}⋮2015\) then   2016^{y + 1} = 1

<=> y + 1 = 0

<=> y = -1 

Change x = 0 , y = -1 into A , we have A = 0  (satisfy)

Conclude : With x = 0 , y = -1 then A is a square number 

Denote \(\left\{{}\begin{matrix}A=\dfrac{2017^x-2016^{y+1}}{2015}\\B=2017^x-2016^{y+1}\\C=2015\end{matrix}\right.\)

For A is a square number so that \(B⋮C\)

<=> \(2017^x-2016^{y+1}⋮2015\)                                                        (1) 

<=> 2017^x - 2^x + 2^x - 2016^{y + 1} + 1 - 1 \(⋮\) 2015

<=> 2^x - 1 \(⋮\) 2015 

=> \(2^x\equiv1\left(mod2015\right)\)     (*)

+) With x = 0 then (*) is true

+) With x = 1 then (*) is false

+) With x = 2 then (*) is false

+) With x \(\ge3\) => \(\left[{}\begin{matrix}x=3k\\x=3k+1\\x=3k+2\end{matrix}\right.\left(k>0\right)\) 

Change three cases into (*) , it's still wrong
So , x = 0 is satisfy 

Because x = 0 => (1) <=> \(1-2016^{y+1}⋮2015\)

For \(1-2016^{y+1}⋮2015\) then 2016^y+1 = 1 => y + 1 = 0 => y = -1

Change x = 0 , y = -1 into A 

A = 0 (satisfy the problem) 

Conclude : With x = 0 , y = -1 then A is a square number