Search Results
| Search type | Search syntax |
|---|---|
| Tags | [tag] |
| Exact | "words here" |
| Author |
user:1234 user:me (yours) |
| Score |
score:3 (3+) score:0 (none) |
| Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
| Views | views:250 |
| Code | code:"if (foo != bar)" |
| Sections |
title:apples body:"apples oranges" |
| URL | url:"*.example.com" |
| Saves | in:saves |
| Status |
closed:yes duplicate:no migrated:no wiki:no |
| Types |
is:question is:answer |
| Exclude |
-[tag] -apples |
| For more details on advanced search visit our help page | |
Results tagged with polynomials
Search options not deleted
user 17907
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.
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 …
- 4,746
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 …
- 4,746
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 …
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 …
- 4,746
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 …
- 4,746