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Questions in which polynomials (single or several variables) play a key role. It is typically important that this tag is combined with other tags; polynomials appear in very different contexts. Please, use at least one of the top-level tags, such as nt.number-theory, co.combinatorics, ac.commutative-algebra, in addition to it. Also, note the more specific tags for some special types of polynomials, e.g., orthogonal-polynomials, symmetric-polynomials.

1 vote

Irreducibility of polynomials over some number fields

Here is a different approach, which is arguably a bit more elementary. If $f=X^n-p$ splits in $K$, and $g$ is one of its factors, then the constant term of $g$, being a product of zeros of $f$, must b …
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1 vote

Why do we make such big deal about the 'unsolvability' of the quintic?

I like this question because I agree with its sentiment. Let me give an additional reason why the insolubility of the quintic is an overrated result in my opinion. I believe that we shouldn't even be …
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3 votes
Accepted

What is the state-of-the-art for solving polynomials systems over fields that are not algebr...

For the reals, I particularly like the book by Sturmfels mentioned by Alexandre Eremenko. For the rational numbers, you can hardly do better than Bjorn Poonen's book Rational Points on Varieties, whic …
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13 votes

Finding $q(x)$ such that $p(q(x))$ is reducible over $\mathbb{Q}[x]$

Note: in this answer, I have inadvertently disregarded your requirement for $q$ to have integral coefficients. I do however prove that a $q$ with rational coefficients does exist, so I will just let t …
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20 votes

$P(x)=P(y)$ has infinitely many integer solutions

these two types of solutions, we get that all solutions are of the form $Q((x+k/2)^2)$, where $k$ is any integer and $Q$ is any polynomial (to be more accurate of course, I should say the subset of all polynomials … Morover I think it should be easy to prove that this description coincides with the set of polynomials of the form $Q(x^2+ax+b)$, with $a$ and $b$ integers and again $Q$ any polynomial, which avoids dealing …
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