My $x$ and $y$ 's

Let there be a positive integer $p$ such that

$(2017+p)^2 = (2017+x)(2017+y)$

Simple rearrangement of terms will give us

$4034p + p^2 = 2017(x+y) + xy$

$p^2 - xy = 2017(x+y-2p) \:\:\: \text{eq. 1}$

Now we know that the expression $2017 | p^2 - xy$ , or that $p^2 - xy = 2017k$ for some integer $k$ .

And $k$ here is

$k= x+y - 2p$

so $p = \frac{x+y-k}{2}$

Now, we want to make sure that $x$ and $y$ are different, so let us express $y$ in terms of $x$ and some other number $q$ , such that $y = x+q$ .

Substituting $y$ onto $p$ yields us

$p= x + \frac{q-k}{2}$

And substituting this to $\:eq. 1$ , we get

$( x + \frac{q-k}{2})^2 = 2017k + x(x+q)$

$(\frac{q-k}{2})^2 = k(2017+x)$

We can now proceed in finding the minimum $x$ and $q$ , and finding $y$ will easily come next.

For an integer $q$ it will be fairly easy to note that the minimum will be achieved when

$\frac{q-k}{2} = \lceil \sqrt{2017k} \rceil$

and that gives us $q$ in terms of $k$ as

$q = 2 \lceil \sqrt{2017k} \rceil +k$

And then $x$ will just follow.

$x = \frac{1}{k}(\frac{q-k}{2})^2 - 2017$

Of course, so should $y$ , as $y= x + q$ .

We now investigate the values of $x$ and $y$ for increasing $k$ .

$\begin{array}{|c|c|c|c|} \hline \ \ k\ \ &\ \ q\ \ &\ \ x\ \ &\ \ y\ \ \\ \hline \hline \ \ 1\ \ &\ \ 91\ \ &\ \ 8\ \ &\ \ 99\ \ \\ \hline \ \ 2\ \ &\ \ 130\ \ &\ \ 31\ \ &\ \ 161\ \ \\ \hline \ \ 3\ \ &\ \ 159\ \ &\ \ 11\ \ &\ \ 170\ \ \\ \hline \ \ 4\ \ &\ \ 184\ \ &\ \ 8\ \ &\ \ 192\ \ \\ \hline \ \ 5\ \ &\ \ 207\ \ &\ \ 23\ \ &\ \ 230\ \ \\ \hline \end{array}$

We see now that $q$ increases with $k$ , so the minimum $q$ , $x$ and $y$ occurs when $k=1$ , such that $x+y = \boxed{107}$ .

( $2017+x)(2017+y)$ = $45^2$ . $46^2$ ,

So that $x=8,y=99$ .

So $x+y$ = 107.