i have a question regarding roots of equation, find all $a$,such that the cubic polynomial $x^3-bx+a=0$ has three integer roots, how can you solve these by using galois theory,what does the reducible polynomials,splitting fields,field extensions have to do with these, explain each of them,in detail,because it serves as a introduction to galois theory, ok for eg take $b$ to be 3 and list all $a$ such that equation has three integer roots

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    $\begingroup$ Galois theory doesn't help you much here. This is really a system of Diophantine equations. $\endgroup$ – Qiaochu Yuan May 15 '11 at 5:58
  • $\begingroup$ but its a cubic polynomial ,it has to do something related to galois group,irreducible things...@Yuan $\endgroup$ – IDOK May 15 '11 at 6:00
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    $\begingroup$ No, it really doesn't. This isn't a Galois-theoretic question. $\endgroup$ – Qiaochu Yuan May 15 '11 at 6:05
  • $\begingroup$ then what do the galois theory actually do,what are splitting fields for,i was thinking that splitting fields has something to do with the roots,@Yuan $\endgroup$ – IDOK May 15 '11 at 7:00
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    $\begingroup$ Not this. Perhaps you should ask about applications of Galois theory as a separate question, although I imagine this has already come up on math.SE at some point. $\endgroup$ – Qiaochu Yuan May 15 '11 at 7:07

Your problem is a classical diophantine problem that Fermat or Newton could have studied, long before Galois.

In the case you mention, you don't really need Galois Theory.

  • You can use symmetric polynomials of the roots to derive constraints on the coefficients. EDIT: (see Quiaochu's answer for a simple and classical way to start doing this)

  • there is a complete algebraic solution of the 3rd degree equation giving the roots in term of the coefficients.

You might enjoy this representation of solutions for small $b$s and $a$s.

grid of integer solutions of $x^3 - b x + a =0$

with red squares indicating combinations of $b$ and $a$ where the polynomial has 3 integer solutions and pink squares where there is only one (in this case the polynomial can be factorized with an integer linear factor, the two other solutions being complex conjugates). You can guess with this image (but need to prove) that there is no solution for negative b, that the solutions are symmetric around the vertical axis and that solutions arise as intersections of straight lines representing one parameter families of equations with at least one integer root.

Galois Theory is not about whether roots of polynomial equations are integers. It has been developed and applied initially to study whether roots can be finitely expressed by addition, substraction, multiplication, division, n-th power and n-th root extraction starting from given quantities, including the coefficients of the polynomials, that is whether these solutions are algebraic in terms of the coefficients and other quantities.

The theory provides among other things a deep and far-reaching understanding of what happens for polynomial equations with degree greater or equal to 5: the individual solutions cannot always be expressed as an algebraic expressions of the coefficients, while some expressions involving simultaneously all the roots can.

Of course, if you can prove with an application of Galois Theory that the roots of a polynomial cannot be expressed algebraically in terms of the coefficients and if the coefficients are integer or rational, then you can deduce that the roots cannot be integers, but this is not what you are asking for. Galois Theory will not give you all the necessary and sufficient conditions on the coefficients for the roots to be integers. For polynomials of degree inferior to 5, Galois Theory is not needed and not sufficient for the result your looking for.

  • $\begingroup$ Galois theory is not really about radical extensions either. That happens to be one of the questions it can address, but it can address many other questions. (Also, since the elementary symmetric polynomials in the roots are just the coefficients, $b$ and $a$ already have to be integers.) $\endgroup$ – Qiaochu Yuan May 15 '11 at 6:18
  • $\begingroup$ @Qiaochu : right, it was developed around that particular problem but is not restricted to it. I will change my wording. $\endgroup$ – ogerard May 15 '11 at 6:28
  • $\begingroup$ @Qiaochu : I am interested to know if you agree with the version. $\endgroup$ – ogerard May 15 '11 at 13:42
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    $\begingroup$ It looks fine. The cubic formula isn't terribly useful for figuring out when the roots are integers, though. $\endgroup$ – Qiaochu Yuan May 15 '11 at 13:58

Suppose the polynomial has three integer roots $r_1, r_2, r_3$. Then $(x - r_1)(x - r_2)(x - r_3) = x^3 - bx + a$, hence

$$r_1 + r_2 + r_3 = 0$$ $$r_1 r_2 + r_2 r_3 + r_3 r_1 = -b$$ $$r_1 r_2 r_3 = -a.$$

Squaring the first equation gives $r_1^2 + r_2^2 + r_3^2 = 2b$, which immediately tells you that for fixed $b$ there are only finitely many possibilities for the roots, and from here it's casework for any fixed $b$.

For example, for $b = 3$ we get $r_1^2 + r_2^2 + r_3^2 = 6$, which has solutions $(\pm 1, \pm 1, \pm 2)$ up to cyclic permutation, and of these solutions only $(-1, -1, 2)$ and $(1, 1, -2)$ add up to zero. Hence the possible values in this case are $a = \pm 2$.

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    $\begingroup$ nice answer sir $\endgroup$ – IDOK May 15 '11 at 16:58

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