# Roots of Polynomial for a Derivative

An elliptic curve can be written as y^2=x^3+ax+b. I need a proof for why x^3+ax+b either have 3 real roots or 1 real root and 2 complex roots. I don't have anything that I know about it prior to asking for help here at Brainmass.

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Problem:

An elliptic curve can be written as y^2=x^3+ax+b. I need a proof for why x^3+ax+b either have 3 real roots or 1 real root and 2 complex roots.

Solution:

It is important first to point out that the coefficients a and b are real, otherwise the given polynomial will have only complex roots. In this case, one needs to analyze the polynomial function

where ( 1)

One first remark: ( 2)

Since f(x) is a continuous function, it means that there exists at least one point for which . In other words, the equation admits always at least one real solution.

The problem now is te nature of the other two solutions of the equation . The following cases are possible:

1) The derivative of f(x)

( 3)

has real roots. In this case, it is necessary that . Let's denote the (real) solutions of equation :

( 4)

If the values of f(x) in those points are of opposite sign:

( 5)

then, the equation will have 3 real solutions.

Otherwise, the equation will have one real solution and two complex solutions.

For the particular case when or , the equation will have 3 real solutions, but one of them will be double ( ).

2) The derivative of f(x)

( 6)

has complex roots. This happens when . In this case, the function f(x) will be monotonic on R (increasing, there are no extrema) and, as a consequence, the equation will have only one real solution.

Let's suppose now that one root of equation is complex. According to one of Viete formulas, we have

( 7)

Since and , it follows that . This happens only when and are complex conjugate:

( 8)

As such, it is not possible for the given polynomial with real coefficients to have only one complex root: it will have either two complex roots (which will be necessarily conjugate), or only real roots.

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