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!
 A: 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. 
A: I am posting this method because first of all, it is extremely elementary (it barely requires anything) and secondly, no one has done it this way yet.
Claim: $gcd(x,y)=1$
Proof: (By contradiction)
Let $gcd(x,y) = d$.
Then, $x = ad, y= bd$, where $a$ and $b$ are co-prime.
The equation now becomes:
$d^3(a^3 - b^3) = d^2ab + 61$.
Case 1: $gcd(d,61) = 1$
Clearly, the left hand side is divisible by $d^3$ whereas, the right hand side is co-prime to $d$. Therefore, equality can never hold. 
Case 2: $gcd(d,61) = 61$ can be handled similarly.
Therefore, either $x^2 + y^2 = 61$ and $x - y = 1$ or the other way round. This is because if $x$ and $y$ are co-prime, $gcd(x^2+xy+y^2, x - y) = 1$ (I hope you can see why). Now, it’s quite obvious to see the only possible solutions. 
EDIT 1: I forgot to write about the case when $xy+61$ can be divided into two co-prime factors. (Maybe, this is the reason I didn’t get the expected positive response). It’s quite easy to handle this one.
So, let’s say $xy+61=\frac{xy+61}{p} \times p$ where $gcd(p, \frac{xy+61}{p}) = 1$ and $p, \frac{xy+61}{p} \neq 1$
Now, WLOG, assume $x-y = p$.
As a result, $px^2 + (p-1)xy + py^2 = 61$.
After re-arranging and setting $x = y+p$, we get, 
$(3p - 1)(y+p)(y) = 61 - p^3$.
LHS, being always positive, limits the range of p to [1, 3]. But, $3p - 1$ doesn’t even divide $61 - p^3$ except for the case when $p = 1$.
Hence, the only possible solution over the positive integers is $(6, 5)$.
QED
A: More generally,
suppose
$x^3-y^3=xy+m
$
where $x, y, m > 0$.
I will show that
$m \ge x^2+y^2$,
$x \ge y+1$,
$x \lt \sqrt{m}$
and
$y
\le \frac12(\sqrt{2m-1}-1)
$.
If $m=61$ then
$x \le 7$
and
$y 
\le \frac12(\sqrt{120}-1)
\approx 4.97
$
so
$y \le 4
$
and
$x \le \sqrt{61-y^2}
$.
We must have
$x > y$
so that
$x \ge y+1$
and
$xy+m
=(x-y)(x^2+xy+y^2)
\ge x^2+xy+y^2
$
so
$m
\ge x^2+y^2
$.
Therefore
$m \gt x^2
$
and
$x \ge y+1$
so that
$m
\ge (y+1)^2+y^2
=2y^2+2y+1
$
or
$2m-1
\ge 4y^2+4y+1
=(2y+1)^2
$
so
$y
\le \frac12(\sqrt{2m-1}-1)
$.
