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Let f, g, and hbe non-constant, relatively prime polynomials such that f + g = h. Then
wheredenotes the number of distinct roots of f.
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 assumingand
, and considering the systems of equations from the coefficients. One set of solutions is
, giving the counterexample:
=f(x)+g(x)=3(\sqrt{3}+3)x^2-3(4\sqrt{3}+6))x+(15\sqrt{3}+27)=)
So these polynomials, being nonconstant and sharing no roots, satisfy the requirements. However,has three distinct roots, thus
which is absurd.
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 havewhere the equality holds if
and
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 calculatedOriginally 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, 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
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