A Novel Bound for Polynomials?

• Jul 9th 2009, 08:15 PM
AlephZero
A Novel Bound for Polynomials?
True or false:

Let f, g, and h $\displaystyle \in\mathbb{C}[x]$ be non-constant, relatively prime polynomials such that f + g = h. Then

$\displaystyle \max (\deg f, \deg g, \deg h) \leq n_0(fgh) - 1,$

where $\displaystyle n_0(f)$ denotes the number of distinct roots of f.
• Jul 10th 2009, 12:42 PM
siclar
Ok, I have not found an elegant way to approach this but with some intuitive guesses I have found a counterexample showing the claim to be false. Now clearly, from being relatively prime and nonconstant, $\displaystyle n_0(fgh)\geq 3$, so the first hope for a counter example would be $\displaystyle \max (\deg f, \deg g, \deg h)=3, n_0(fgh) =3,$ in which case the inequality fails. For this to work, each polynomial must be either a perfect square or perfect cube of a linear term. It is not difficult to show that we cannot sum two perfect cubes to another perfect cube by considering the resulting system of equations of the coefficients.

My solution was constructed by assuming $\displaystyle f(x)=(x-a)^3, g(x)=(b-x)^3$ and $\displaystyle f(x)+g(x)=h(x)=w(x-c)^2$, and considering the systems of equations from the coefficients. One set of solutions is $\displaystyle a=1, b=\sqrt{3}+2, w=3(\sqrt{3}+3), c=-\sqrt{3}-1$, giving the counterexample:

$\displaystyle f(x)=(x+1)^3=x^3+3x^2+3x+1$
$\displaystyle g(x)=(\sqrt{3}+2-x)^3=-x^3+3(\sqrt{3}+2)x^2-3(4\sqrt{3}+7))x+(15\sqrt{3}+26)$

$\displaystyle h(x)=f(x)+g(x)=3(\sqrt{3}+3)x^2-3(4\sqrt{3}+6))x+(15\sqrt{3}+27)=$$\displaystyle 3(\sqrt{3}+3)*(x^2-(\sqrt{3}+1)x+(\sqrt{3}+2))=3(\sqrt{3}+3)(x-(\sqrt{3}+1)/2)^2$

So these polynomials, being nonconstant and sharing no roots, satisfy the requirements. However, $\displaystyle (fgh)(x)=3(\sqrt{3}+3)(x-(\sqrt{3}+1)/2)^2(x+1)^3(\sqrt{3}+2-x)^3$ has three distinct roots, thus $\displaystyle 3=\max (\deg f, \deg g, \deg h)=\leq n_0(fgh)-1=3-1=2$ which is absurd.
• Jul 10th 2009, 01:23 PM
AlephZero
Without giving away the answer to the OP, I will say that I am afraid your argument does not work, siclar, though it was a noble effort!

In general, we have $\displaystyle n_0(fg)\leq n_0(f) + n_0(g),$ where the equality holds if $\displaystyle f$ and $\displaystyle g$ are relatively prime. So I think your initial step was the problem...
• Jul 10th 2009, 02:08 PM
NonCommAlg
Quote:

Originally Posted by AlephZero
True or false:

Let f, g, and h $\displaystyle \in\mathbb{C}[x]$ be non-constant, relatively prime polynomials such that f + g = h. Then

$\displaystyle \max (\deg f, \deg g, \deg h) \leq n_0(fgh) - 1,$

where $\displaystyle n_0(f)$ denotes the number of distinct roots of f.

see Mason-Stothers theorem.
• Jul 10th 2009, 02:27 PM
siclar
Quote:

Originally Posted by AlephZero
Without giving away the answer to the OP, I will say that I am afraid your argument does not work, siclar, though it was a noble effort!

In general, we have $\displaystyle n_0(fg)\leq n_0(f) + n_0(g),$ where the equality holds if $\displaystyle f$ and $\displaystyle g$ are relatively prime. So I think your initial step was the problem...

Hm, I'm afraid I don't see how this changes anything I wrote, seeing as I explicitly calculated $\displaystyle n_0$, or rather constructed my polynomials to give it a specific value.

Having googled NCA's suggestion, I see I must be wrong but I can't figure out where my logic falls apart. I double checked my arithmetic on the polynomial computations and that all seems fine