True or false:

Let f, g, and hbe non-constant, relatively prime polynomials such that f + g = h. Then

wheredenotes the number of distinct roots of f.

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- July 9th 2009, 09:15 PMAlephZeroA Novel Bound for Polynomials?
True or false:

*Let f, g, and h**be non-constant, relatively prime polynomials such that f + g = h. Then*

*where**denotes the number of distinct roots of f.* - July 10th 2009, 01:42 PMsiclar
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, , so the first hope for a counter example would be 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 and , and considering the systems of equations from the coefficients. One set of solutions is , giving the counterexample:

So these polynomials, being nonconstant and sharing no roots, satisfy the requirements. However, has three distinct roots, thus which is absurd. - July 10th 2009, 02:23 PMAlephZero
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 where the equality holds if and are relatively prime. So I think your initial step was the problem... - July 10th 2009, 03:08 PMNonCommAlg
- July 10th 2009, 03:27 PMsiclar
Hm, I'm afraid I don't see how this changes anything I wrote, seeing as I explicitly calculated , 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