A cubic integer polynomial must have an irrational root.

irrational-numberspolynomialsroots

Prove that a cubic integer polynomial $ax^3+bx^2+cx+d$ where $ad$ is odd and $bc$ is even must have an irrational root.

The above problem is from the journal The Mathematics Student of The Indian Mathematical Society.

My approach:

Consider the roots are $x_1, x_2, x_3.$ Since the polynomial is cubic, only 3 cases are possible:

  1. $3$ distinct real roots.
  2. $3$ real roots where $1$ root has multiplicity $2$.
  3. $1$ real and $2$ complex conjugate roots.

Since we want an irrational root, I could eliminate the last $2$ possibilities.

Also, we have the following relation

$$ x_1+x_2+x_3 = \frac{-b}{a}$$
$$ x_1x_2+x_2x_3+x_3x_1 = \frac{c}{a}$$
$$ x_1x_2x_3 = \frac{-d}{a}$$

Since $ad$ is odd, both $a$ and $d$ are odd and $bc$ is even implies at least one of $b$ and $c$ is even.

How do I proceed now?
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Best Answer

I will address the remaining case #1 in your question.

Assume that such a polynomial has 3 real roots, and that the factorization is $(Ax+X)(Bx+Y)(Cx+Z)$. First, since $a=ABC$ is odd, then $A$, $B$, and $C$ are all odd. Then, since $d=XYZ$ is odd, then $X,Y,Z$ are also all odd.

Reduce the above factorization (mod 2), and we get $(x+1)(x+1)(x+1)$. This multiplies out to $x^3+3x^2+3x+1$. This means that the $b$ and $c$ factors are also odd, contradicting the assumption that their product is even. So, this is not possible.

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