Integral solutions to the diophantine equation $y^2=x^3+2017$. Case 1: $y^2=0 \mod{3}$
\begin{align*}
& y^2 = 3b \quad (b=3k^2) \implies x^3=3a +2  \\
& 3b = 3a + 2 + 2017 \implies b-a = 673 \\
& k^2 = \frac{a-2}{3} + 225 \\
\end{align*}
Then I just took $a =2$ since 225 is already a perfect square. Thus I got the solution $(x,y) = (2,\pm 45)$. This is the only solution I was able to get.
I tried to do the second case, i.e, $y^2 \equiv 1 \mod 3$ as I did Case 1 but did not have any luck. Here are my proceedings.
Let $y^2 = 3b + 1$ and $x^3 = 3a$ where $a = 9k^3$
Then doing the same thing above we get,
\begin{equation*}
k^3 = \frac{b+3}{9} - 75
\end{equation*}
Now I am stuck, since I cannot find any solutions to the above equation.

Comment:
You can try this method:
Take floor of $2017^{\frac 13}=12$
You can check 12 and less to see it gives a solution:
$2017=12^3+17^2=y^2-x^3$
Which gives $y=\pm 17$ and $x=-12$
Among other numbers bellow 12 only 2 gives integer for y:
$2017=45^2-2^3=y^2-x^3$
Which give what you found; $x=2$ and $y=\pm 45$
This method is for small numbers.
