Is there a general method to determine the [non-]solutions to $ax^4+bx^2+c=y^2$? I am looking for general methods to solve, in integers $x$ and $y$, the equation
$$
ax^4+bx^2+c=y^2
$$
where $a,b,c$ are given integers, and by "solve" I mean:
(i) show there are no [non-trivial] solutions; or
(ii) give a complete finite solution; or
(iii) provide a parameterization, recursion, or algorithm by which all solutions may be obtained.
As an example, I believe I have recently solved the case $(a,b,c)=(48,12,1)$, i.e., I think I have shown, using ad hoc elementary methods, that the equation $$48x^4+12x^2+1=y^2$$ has only the trivial solution $(x,y)=(0,\pm 1)$. [n.b. I could post my proof here, but it is ultimately irrelevant to my present question.]
Now I'm wondering if this particular wheel has already been invented.
 A: A general and elementary method for these equations is given in the article Fermat and the difference of two squares, The Mathematical Gazette, Vol96, Number 537,Nov 2012, pp.480-491.
These equations can have a small set of easily determined 'trivial' solutions together with one or more infinite sequences of solutions derived from base solutions with neither $a$ nor $c$ squares.
Although this is proved by elementary means, involving no more than the difference of two squares and Fermat's proof by infinite descent, it is interesting to see the parallels with the Torsion group and Rank of an elliptic curve solution.
The connection with elliptic curves is explored further in the same journal, New points from old, The Mathematical Gazette, Vol96, Number 537, Nov 2012, pp. 492-498, where a group structure is set up directly on these curves without the usual process of conversion into a standard elliptic curve form. (The method described there extends easily to curves with $y^2=ax^4+bx^2+c$ and, providing one knows a rational point, extends to general quartics).
