Please allow me to continue being a skeptic. I am having some mental block.
Why is it that if , are roots of , then ? The converse is obviously true. I mean that we are completely used to this, but since the question is so basic, the proof should be similarly low-level, based on simple arithmetic facts.
Because the set of all polynomials, p(x), of degree two or less, such that and , forms a one-dimensional subspace of the vector space of all polynomials of degree two or less. And that is, in my opinion, based on very "low level arthmetic facts"- it's just tedious to say without using the "vector space" terminology.
Let's consider only polynomials of degree 2 or less.
This says that there exists a polynomials with the following property. For every , iff there exists an such that .
This fact it pretty similar to the original claim that iff . The question is why it holds.
The temptation is precisely to resort to more complex facts, but it is not clear that this does not involve circular reasoning. I would like to see a proof from simpler facts.
Here is one way, inspired by Little Bézout's theorem. Suppose and and are the only roots of . Then, one can check directly that
Indeed, the right-hand side expands to , but since is a root, .
The roots of the left-hand side are and , and the roots of the right-hand side are and . Therefore, .
Thank you, emakarov, HallsofIvy, and Ithaka/
I have another question.
I was able to solve the problem;
In the above equation, solve when
When or
When or
I don't understand why in both circumstances ( and )
I know you can say that because the problem says "when ", but I'm not satisfied with that answer.
I think I may be content if I do the same problem with a different way of approaching it...