Is Pell Relevant Here?

From the equality: $2015a^2+a=2016b^2+b$ $(1)$

We can easily find $a \ge b$ .

If $a=b$ then $a=b=0$ hence $\sqrt{|a-b|}=0$ is an integer.

If $a>b$ , we can write $(1)$ as:

$b^2=2015(a^2-b^2)+(a-b)$

$\Rightarrow$ $b^2=(a-b)[2015(a+b)+1]$ $\Leftrightarrow$ $b^2=(a-b)(2015a+2015b+1)$ $(2)$

Let $(a,b)=d$ then $a=md$ , $b=nd$ , where $(m,n)=1$ . Since $a>b$ then $m>n$ , and put $m-n=p>0$ .

Let $(p,n)=x$ then $p$ is divisible by $x$ , $n$ is divisible by $x$ , and also $m$ is divisible by $x$ . That follows $x=1 \Rightarrow (p,n)=1$ , and $a-b=pd$ .

Putting $b=nd$ , $a-b=pd$ in $(2)$ , we will get

$n^2d=p(2015dp+4030dn+1)$ $(3)$

From $(3)$ we get $n^2d$ is divisible by $p$ and with $(p,n)=1$ , it follows $d$ is divisible by p.

Also from $(3)$ we will get $n^2d=2015dp^2+4030dnp+p$ and then $p=n^2d-2015dp^2-4030dnp$

Hence $p=d(n^2-2015p^2-4030np)$ , i.e. $p$ is divisible by $d$ $\Rightarrow p=d$ and then $a-b=pd=d^2$ and $\sqrt{|a-b|}=d$ is an integer.