Rational solutions to $x^4+y^4=cz^2$ Suppose $c\neq 1$ is a squarefree number, and consider the curve $x^4+y^4=cz^2$. How can I find rational points on this curve?
What I really want to know is how to transform this into an elliptic curve. I'm confused for the following reason: this curve has three variables, while elliptic curves have two variables.
 A: From,
$$x^4+y^4=cz^2\tag1$$
If you do the following change of variables, 
$$X = -\frac{4 x^{2} y^{2}}{z^{2}}\tag2$$ 
$$Y = \frac{4 xy(x^{4} - y^{4})}{z^{3}}\tag3$$ 
then you can check that,
$$Y^2 = X^3 - 4c^2X\tag4$$ 
which is an elliptic curve in Weierstrass form, and is famously related to congruent numbers.
A: I am confused too, by the same reasons you said . Anyway, I give you here two comments (not exactly an answer). 
1) The diophantine equation $ax^4+by^4=cz^2; (x, y)=1$ has a long history and there is a lot of cases of impossibility that have been established (Nagell among others) so take $c\neq 1$ could be too general. Another thing is to calculate $a^4+b^4$, say equal to n, so you get a solution (a, b,1) if n is squarefree or a better solution if does not. (of either, $x^4+y^4=nz^2$ or $x^4+y^4=cz^2$ where c divides n).
2)(Concerning your goal to find out an elliptic curve that works) It is known the quartic $$F(x,y)=ax^4+4bx^3y+6cx^2y^2+4dxy^3+ey^4$$ has as invariants $g_2=ae-4bd+3c^2$ and
$g_3$=det$\begin{pmatrix} a & b& c \\ b&c &d \\c &d &e  \\ \end{pmatrix}$, and the quartic 
$$y^2=x^4+6cx^2+4dx+e$$  (where you have divided $F(x,y)$ by $y^4$ and maked $a=1;b=0)$) is birational equivalent to the cubic $$t^2=4s^3-g_2s-g_3$$ via the function
$(2x(s+c), y)\to (t-d, 2s-x^2-c)$.
Searching of applying this, $x^4+y^4$ has invariants $g_2=g_3=0$ and this curve would be equivalent to the cubic $t^2=4s^3$ which clearly is not elliptic.
I repeat, I am confused too and I give you just a hint maybe worthless here, just wishing  to assist you in your question.
