Solve $x^2 + 12 = y^4$ in the integers. 
Solve $x^2 + 12 = y^4$ in the integers.

My thought was to factor $x^2 - y^4$ = $(x - y^2)(x + y^2) = -12$. I'm not sure if this is the right approach, and I'm not sure where to go from here. 

Rearrange as
$$y^4-x^2=12$$
Because they are squares, any positive value of $y$ or $x$ has a corresponding negative value, so it’s sufficient to find the positive values.
$$(y^2-x)(y^2+x)=12$$
So $[(y^2-x),(y^2+x)]$ is $(1,12)$ or $(2,6)$ or $(3,4)$
(Note: because of the symmetry, we can ignore $(12,1)$ or $(6,2)$ and $(4,3)$.)
Put $a=(y^2-x)$ and $b=(y^2+x)$ then $y^2=(a+b)/2$ and $x=(b-a)/2$
Clearly only $(2,6)$ can fit, as the others give fractions.
So $y^2=(2+6)/2=4$ and $x=(6-2)/2=2$ 
The positive answer is $(x,y)=(2,2)$ and the full answer is
$$(x,y)=(\pm2,\pm2)$$
Please let me know if you need any more details.
