Proving a cubic polynomial has no rational roots

Question

This is an exercise in polynomials/algebra/discrete mathematics I have just met:

For odd integers $a,b \in \mathbb{Z}$ we are asked to show the polynomial $ x^3+ax+b$ has no rational roots.

The first thing that came to my mind is the rational root test which says that for a rational root $x=\frac{p}{q}$ written as a reduced fraction where $p,q \in \mathbb{Z}$ with $ q\neq 0$, then $q$ is a factor of 1 and $p$ is a factor of $b$ which means that a rational root must actually be an odd integer (positive or negative) factor of $b$, but I am stuck here. All help appreciated.

score 1 · Accepted Answer · answered Aug 24 '20 at 20:34

1

You've got the right idea. Your denominator $q$ should be $\pm 1$, so your rational root must actually be an integer. As a result, you just need to show that $$x^3+ax+b\neq 0$$ for all integers $x,a,b$ where $a$ and $b$ are odd. Since you're given that $a$ and $b$ are odd, it makes sense to look at this polynomial modulo $2$; what do you see?

answered Aug 24 '20 at 20:34

Carl Schildkraut

32,931

actually since $b$ is odd and $x$ is an integer factor of it can we also deduce that $x$ must be odd? – Croc2Alpha Aug 24 '20 at 20:37
1

@kroner Yes, exactly, although that isn't strictly necessary. – Carl Schildkraut Aug 24 '20 at 20:44
thank you very much – Croc2Alpha Aug 24 '20 at 20:45
Please search to help avoid adding more dupe Q&A's to dupes of FAQs. – Bill Dubuque Aug 24 '20 at 20:56

score 0 · Answer 2 · answered Aug 24 '20 at 20:40

0

If the root $r$ is odd, $r^3+ar+b$ is an odd number, a contradiction.

answered Aug 24 '20 at 20:40

Proving a cubic polynomial has no rational roots

2 Answers2