Diophantine equations with cubic roots. 
Find all positive, integral $x, y$ such that 
  $$x^3-y^3=xy+61$$

Little work:
$$(x-y)(x^2+xy+y^2)-xy=61$$
Also $x^3-xy-y^3$ seems to to be unable to be "completed by the square," or factored, so I'm stuck. Thanks!

Draw some pictures. There is an integral point on it $(6,5)$
As soon as real $x > 6,$ we get
$$ x-1 < y < x  $$
so that $y$ cannot be an integer when $x > 6$
Tuesday:
We know that $x > y$ are positive. So,
$(x-y)^2 > 0$ or $x^2 - 2xy + y^2 > 0,$ or $x^2 + y^2 > 2xy.$ Add $xy$ to both sides with positive $x,y$ we get
$$x^2 + xy + y^2 > 3xy > 0$$ and
$$   \frac{xy}{x^2 + xy + y^2} < \frac{1}{3}. $$
Since $$ (x-y)(x^2 + xy + y^2) = xy + 61,  $$ we have
$$ x-y = \frac{xy}{x^2 + xy + y^2} + \frac{61}{x^2 + xy + y^2} < \frac{1}{3} + \frac{61}{x^2 } $$
since $x^2 < x^2 + xy + y^2$ when $x,y>0.$
$$ \color{magenta}{ x-y < \frac{1}{3} + \frac{61}{x^2 }} $$ 
Therefore, when $x \geq 10$ is an integer, we find $\frac{61}{x^2} < \frac{2}{3} \; , \;$ so that $0 <x-y < 1$ 
and $y$ is not an integer.
We need only check $x$ values up to $9,$ and $x=6,y=5$ is the only one that works.
There are just the two integer points, but infinitely many rational points. The tangent line at $(-5,-6)$ intersects the curve in a third point, that being
$$  \left(  \frac{1385}{341}, \frac{384}{341} \right)  $$
Given a rational point on the curve, the tangent line there meets the curve in another rational point. Given two distinct rational points on the curve, the line through them meets the curve in a third rational point. 
