<s>[INST] 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 $(x, y) = (3, 2)$
 [/INST] COMMENT.-You have solved easily the problem algebraically and you want to know how solve it in number theory. Well your curve is an elliptic curve of rank $1$ whose generator is the point $(3,2)$ which is the only point of integer coordinates and it is determined using your algebraic way. For the rest it has an infinity of rational points whose set is dense in the curve and that can be calculated by the method of chords and tangents. That is what number theory in connection with elliptic curves  tells us about this problem (algebraic number theory deals with the other non-rational  algebraic points, i.e. non trascendental, of the curve).
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