calculate the Integer solutions of an equation

Question

How does anyone calculate the "integer solutions" of an equation in this example:

$$ x(x+1) + 62 = y^2 $$

https://www.wolframalpha.com/input/?i=x*%28x%2B1%29%2B62%3Dy%5E2

here the Integer solutions are :
$x = -62$, $y= 62$
$x = -2$, $y= 8$
$x = 1$, $y= 8$
$x = 61$, $y= 62$

How do they get $8$?

Answer 1

Hint

The equation can be written as $$(2x+1)^2-(2y)^2=-247 \implies (2x+1-2y)(2x+1+2y)=-247.$$ Note that $247=13 \cdot 19$. So factor pairs are $(1,247), \, (13,19)$.

Answer 2

Multiply it by 4: $$4x^2+4x+248=4y^2$$ so $$(2x+1)^2+247 = 4y^2$$ and thus $$4y^2-(2x+1)^2=247$$

Can you end it now?