Finding rational points on a non-trivial algebraic curve The curve in question is defined as the set of all points $(x,y)$ where both $x$ and $y$ are in $(0,1)$, and that satisfy the following:
$$ \left( x^4 - 6 x^2 + 1 \right)  \left( y^4-6y^2+1
 \right) -64 x^2 y^2 = 0$$
I do not know how I could identify even a single rational point on this curve, let alone others.
Any assistance would be appreciated.
 A: The equation
$$ 
\left( x^4-6x^2+1 \right)  \left( y^4-6y^2+1 \right) - 64 x^2 y^2 = 0
$$
has no rational solutions; indeed the only solutions for which both
$X := x^2$ and $Y := y^2$ are rational are those with $x^2 = y^2 = -1$.
Bruno Joyal already noted that $X$ and $Y$ satisfy
$$ 
\left( X^2 - 6X + 1 \right) \left( Y^2 - 6Y + 1 \right) - 64 X Y = 0,
$$
and noted that this equation still has degree $4$ and that
in general a degree-$4$ curve has genus $3$.  The good news is that
in fact this curve has genus $0$.  The bad news is that it is not 
a rational curve over $\bf Q$, and the only rational solution
is the singularity at $(X,Y) = (-1,-1)$.
For the "good news", note that the curve has degree $2$ in each variable
separately, which makes its genus at most $1$; the node at $(-1,-1)$
then drops the genus to zero.  
For the "bad news", compute the discriminant with respect to $Y$,
finding that $8 (X+1)^2 (X^2+10X+1)$ must be a square.  If $X=-1$ then
the equation $(X^2 - 6X + 1) (Y^2 - 6Y + 1) - 64 X Y = 0$ becomes
$8(Y+1)^2 = 0$, so we recover $(X,Y) = (-1,-1)$.
If $X \neq -1$ then $2(X^2 + 10 X + 1)$ must be a square.  But 
$2(X^2 + 10 X + 1) = 2 (X+5)^2 - 3 \cdot 4^2$, and
$2a^2 - 3b^2$ cannot be a square unless $a=b=0$,
because $2$ is not a square mod $3$.
[We could also show this using congruences modulo $8$.]
Hence $(-1,-1)$ is the unique rational solution of
$(X^2 - 6X + 1) (Y^2 - 6Y + 1) = 64 X Y$, and we're done.
A: First, you might want to try finding a rational point on 
$$(X^2-6X+1)(Y^2-6Y+1) = 64XY.$$
It is an easier problem. Then you can try to find a point $(X_0, Y_0)$ for which both $X_0$ and $Y_0$ are squares.
The projective closure $X \subseteq \mathbf P^2$ of the curve $(X^2-6X+1)(Y^2-6Y+1) = 64XY$ is a plane curve of degree $4$. The first thing you should do is check whether it is smooth. I didn't check.
If it is smooth, then it has genus $(4-1)(4-2)/2 = 3$, and Faltings' theorem implies that there are finitely many rational points on $X$. It's very hard to know a priori how many points there are, so if you do a brute force search for points, you might not be able to tell when you have found all of the points.
If $X$ is not smooth, compute the normalization $\widetilde X \to X$. (You can do this completely mechanically, by blowing up at the singularities until they resolve. This is a nice thing with curves that doesn't work in higher dimensions.) Then try to find a rational point on $\widetilde X$. Whether there is much hope will ultimately depend on the genus $g$ of $\widetilde X$. If $g=0$ then you will be able to parametrize $X$ using rational functions (over $\overline{\mathbf Q})$. If $g=1$ then $\widetilde{X}$ becomes an elliptic curve after a finite base change and a choice of point (over the new base). In this case, there are algorithms to compute a basis for the group of points. If $g>1$, you'll have to be creative.
Finding rational points on algebraic varieties is, in general, a very difficult problem with no straightforward solution.
