solving cubic diophantine equation Can someone show me how to find all solutions in positive integers to the diophantine equation: $$x^3 + y^3 = 35$$ I know how to do it algebraically, but I want to know how you solve it in number theory.
My Algebraic Approach:
$x^3 + y^3 = (x + y)\cdot (x^2 - xy + y^2) = 35$.
The only integer factors of $35$ are $(1, 35)$ or $(5, 7)$. There are no integers $x$ and $y$ that add to $1$, so $x + y = 5$ or $7$.
Using $x + y = 5$ we get that $y = 5 - x$, so $y^3 = 125 - 75x + 15x^2 - x^3$ so
$x^3 + y^3 = 15x^2 - 75x + 125 = 35$
or
$$15x^2 - 75x + 90 = 0$$
$$x^2 - 5x + 6 = 0$$
$$(x - 3)(x - 2) = 0$$
So $x = 2$ or $x = 3$
Thus $y = 3$ or $y = 2$ respectively.
Using $x + y = 7$ we get that $y = 7 - x$, so $y^3 = 343 - 147x + 21x^2 - x^3$ so
$$x^3 + y^3 = 21x^2 - 147x + 343 = 35$$
$$21x^2 - 147x + 308 = 0$$
$$3x^2 - 21x + 44 = 0$$
or
$$x = \frac{21 \pm \sqrt{-87}}{6}$$
Since x is complex, this can't be a solution.
So $(x, y) = (2, 3)$ or $(x, y) = (3, 2)$

To bound all solutions
in integers
(not just to positive integers)
to
$x^3+y^3 = n$:
First,
using the factorization
$x^3+y^3
=(x+y)(x^2-xy+y^2)
$,
we get
possible values
for $x+y$
since $(x+y) | n$.
Then,
since
$x^2-xy+y^2
=(x+y)^2-3xy
$,
we get possible values
for $xy$
($xy
=\dfrac{(x+y)^2-n/(x+y)}{3}
$)
and
this gives $x$ and $y$.
Note (added after a comment):
If $a=x+y$ and $b=xy$ are known,
then,
since
$(r-x)(r-y)
=r^2-r(x+y)+xy
=r^2-ar+b
$,
$x$ and $y$
are the roots of
$r^2-ar+b$.
So use the quadratic formula:
$r
=\dfrac{a\pm \sqrt{a^2-4b}}{2}
$.
(Additional stuff to make the solution explicit)
We have
$a = x+y$
 (the divisor of $n$), 
$b = xy = ((x+y)^2−n/(x+y))/3=(a^2−n/a)/3$. 
Then the discriminant
$d
=a^2-4b
=a^2-4(a^2−n/a)/3
=(4n/a-a^2)/3
=(4n-a^3)/(3a)
$
and the roots are
$\dfrac{a\pm \sqrt{(4n-a^3)/(3a)}}{2}$.
If we adopt the convention
that $x < y$,
then
$x = \dfrac{a- \sqrt{(4n-a^3)/(3a)}}{2}$
and
$y = \dfrac{a + \sqrt{(4n-a^3)/(3a)}}{2}$
.
For this to work,
$(4n-a^3)/(3a)$
must be a perfect square.
Some elementary bounds on $x$ and $y$:
If
$0 < x \le y$,
$\sqrt[3]{n/2} < y \lt \sqrt[3]{n}
$.
If $x < 0 < y$,
since
$x^2-xy+y^2
=x^2-xy+(y/2)^2+3y^2/4
=(x-y/2)^2+3y^2/4
> 3y^2/4
$,
$x+y > 0$
or
$y > -x$.
This can be improved to
$x^2-xy+y^2
=x(x-y)+y^2
=(-x)(y-x)+y^2
> y^2
$
and
$x^2-xy+y^2
<3y^2
$.
For every divisor $d$ of $n$,
$d y^2 < n < 3dy^2$,
so
$\sqrt{n/(3d)} < y < \sqrt{n/d}$.
