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The solution of the cubic equation is known in terms of a rational function of a cube root of a square root. If we just want to know the value it's easy to approximate it using a numerical method. I would like to know if the actual form (in terms of roots) of the solution of the cubic has any applications in number theory?

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Solving equations numerically is usually good enough for engineers, say, and they probably don't need the explicit formula very often. Pure mathematicians are usually more interested in understanding how things work rather than computing specific numbers.

There is a lot of information about solutions of a cubic that you cannot read off from approximate solutions. The most obvious such property is their field of definition. E.g. you cannot tell whether a cubic has solutions over the rationals by looking at approximations. Finding rational and integral solutions to polynomial equations is actually a huge part of what algebraic number theory is about in the first place. This area is called the theory of Diophantine equations and goes back to Diophantus of Alexandria, almost 2000 years back. It is usually not thought of as a means to an end, but a goal in its own right, since it is ultimately a question about how natural and rational numbers behave. The latter are some of the most immediate mathematical entities and our desire to understand them is just a manifestation of our curiosity.

To understand Diophantine equations, one often needs to understand not just the field of rational numbers, but more general number fields. So, we are not only interested in whether or not a cubic has a rational solution, but we want to know exactly the field of definition of these solutions, and for that, an algebraic expression for its roots is very handy.

This was a long way of saying that you shouldn't think of the algebraic formula for the roots of a cubic as a means to an end, but rather as an accomplishment on the way to a more general aim: solving Diophantine equations and understanding how integers and the rational numbers work. This more general aim is pursued by number theorists because they find it exciting, not because it has applications.

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Any Diophantine equations which we can solve using this:… ? – quanta Jan 12 '11 at 10:32
@quanta I am afraid, I don't understand what you mean by "this way", since the post you linked to doesn't contain a method for solving Diophantine equations. I should add that nowadays, there exist very sophisticated techniques for solving classes of Diophantine equations or for determining, how many rational solutions an equation has, and there is also a very beautiful and rich conjectural picture available, which appears to be extremely hard to prove. – Alex B. Jan 12 '11 at 12:58

There are probably more interesting and diverse answers, but one immediate application to Galois theory/algebraic number theory is that knowing the solution of the general cubic equation $x^3+ax^2+bx+c=0$ (where the cubic is irreducible over the rationals) allows us to compute the Galois group of a cubic. In particular, the discriminant $D=18abc+a^2b^2−4b^3−4a^3c−27c^2$, which appears under the square root that you mentioned, completely determines the Galois group. In general, the splitting field for the cubic is $K=\mathbb{Q}(r, \sqrt{D})$, where $r$ is any one root of the cubic. If $\sqrt{D}$ rational, then the Galois group is isomorphic to $A_3$. Otherwise, the Galois group is isomorphic to $S_3$.

EDIT: As Qiaochu pointed out, the solution of the cubic is better viewed as an application of Galois theory than the other way around. In general, the particulars of solving polynomial equations (for example, what the solutions look like, i.e. whether we can get them by adjoining roots, etc.) is a motivating problem for a branch of algebra/number theory, and Galois theory provides information with regard to these things. There are a few reasons why the problem of understanding solutions of polynomial equations is interesting. One of them is that this is one of the simplest interesting questions that we can ask in number theory. Another is that polynomials tend to come up in a lot of places, so knowing about their solutions is useful to everyone, not just number theorists. I know I haven't answered your question about applications of the solution of the cubic to number theory (no significant applications come to mind, but maybe there are some), but there is an interesting story here that is motivated by wanting to know about solving polynomial equations. In particular, you might be interested to read about the insolvability of the general quintic by adjoining radicals and how this fits into the history of Galois theory.

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You can prove that the discriminant completely determines the Galois group without knowing the cubic formula. In general it seems easier to me to use Galois theory than to use the cubic formula; if anything I would count the cubic formula as an application of Galois theory instead of the other way around. – Qiaochu Yuan Jan 5 '11 at 20:19
The cubic formula is tedious to derive, so computing the Galois group through Galois theory is quicker. But once one has the cubic equation, it shouldn't be more than some basic algebra to prove the above facts about the Galois group. The fact that the general cubic is solvable by radicals follows from the Galois theory, but is there a derivation of the cubic formula that doesn't go through the elementary methods? – Vitaly Lorman Jan 5 '11 at 20:56
I looked it up and there is. I think you're right that Galois theory more naturally applies to this stuff than the other way around. – Vitaly Lorman Jan 5 '11 at 21:16
In that case I would like to extend my question to include "galois group of a cubic" (I don't know what you could use that for) - thank you! – quanta Jan 6 '11 at 7:23
@quanta: special cases of what you can use Galois theory in general for. What kind of answers are you looking for? – Qiaochu Yuan Jan 11 '11 at 21:37

This doesn't answer the question as stated but instead is a related result.

Siegel's Theorem is an example of using the discriminant of the cubic equation in algebraic number theory. It states that a smooth algebraic curve $C$ of genus $g > 0$ defined over a number field $K$, written in a given coordinate system over affine space, contains only finitely many points with coordinates in the ring of integers of $K$. For example, the elliptic curve $y^{2} = x^{3} + a x + b$ with $a,b,c \in \mathbb{Z}$ and $4 a^{3} + 27 b^{2} \neq 0$ has only finitely many solutions $x , y \in \mathbb{Z}$.

This particular example can be better understood if one knows one solution of the reduced cubic, $x^{3} + p x = q$, which is \begin{eqnarray} x = \sqrt[3]{ \sqrt{D} + \frac{q}{2}} - \sqrt[3]{ \sqrt{D} - \frac{q}{2}}, \end{eqnarray} where $D = \frac{q^{2}}{4} + \frac{p^{3}}{27}$ is the discriminant up to a constant factor (which we define as $\Delta$ below). More generally, the discriminant of the cubic $ax^{3} + bx^{2} + cx + d = 0$ is \begin{eqnarray} \Delta = b^{2} c^{2} - 4ac^{3} - 4 b^{3} d - 27 a^{2} d^{2} + 18 abcd. \end{eqnarray} There are ways to calculate the discriminant without solving the cubic, however. Note that if $\Delta = 0$, then the cubic has repeated roots.

Considering the right side of the defining equation of elliptic curve $C$, we have \begin{eqnarray} \Delta(C) = - (4 a^{3} + 27 b^{2}), \end{eqnarray} which is up to sign the conditional factor above. A nice exercise is to think about what occurs (or is violated) in the case that the associated discriminant vanishes.

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What do you mean by "This result can only be stated if one knows one solution"? Since the result can be stated for curves of arbitrary genus, it clearly doesn't rely on knowing any solutions to any polynomial. Nor do you need Cardano's formula for calculating the discriminant. Again, the discriminant can be calculated for polynomials of arbitrary degree, where there is no Cardano's formula. Maybe I completely misunderstood what you meant? – Alex B. Jan 14 '11 at 15:37
Yes, you misunderstood, but it was my fault. Thanks for pointing it out! Edited. – user02138 Jan 14 '11 at 18:51
Well, I still don't see how your example is an application of Cardano's formula. Nobody calculates discriminants by calculating the roots first. The order in which things were discovered is beside the point. That would be the same as saying that one needs to solve a cubic to compute a cohomology group, because Cardano's formula was discovered before cohomology. And I still don't understand what you mean by "example can be stated if...". Never mind. – Alex B. Jan 15 '11 at 0:39
I agree with Alex; I don't see what this has to do with Cardano's formula and would appreciate that you attempt to communicate this clearly instead of being mysterious and pompous. – Qiaochu Yuan Jan 17 '11 at 5:42
I still agree with Alex. – Qiaochu Yuan Jan 18 '11 at 0:48

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