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 $(x, y) = (3, 2)$

You know that $x$ and $y$ cannot both be greater than or equal to $3$, because this would give you too large of a sum. And $x$ and $y$ must both be less than $4$ by similar reasoning. So in this case, that's probably the best way to approach the problem: Just try $1,2,3$ for $x$ and $y$, where they are not both $3$. And note you can assume $x \leq y$ by symmetry, and modify your answer after the fact based on symmetry and the solutions you get.
