Finding all the integer solutions for :$y^2=x^6+17$ Assume that $x,y$ are integers .How to find the solutions for:
$$y^2=x^6+17$$ 

$y^2-x^6=17\Rightarrow (y-x^3)(y+x^3)=17$.And we know $17$ is a prime.
So the only possibility is  
$y+x^3=17$ and $y-x^3=1$ .
or $y+x^3=-17$ and $y-x^3=-1$
or $y+x^3=1$ and $y-x^3=17$
(As $x,y$ are integers so $(y-x^3)(y+x^3)=17$ implies that the only factors of $17 $ are $y+x^3$ and $y-x^3$ )
$\Rightarrow y=9,x=2$
or $y=9,x=-2$
or $x=-9,y=2$
or $x=-9,y=-2$
