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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.

9 votes
1 answer
990 views

Are polynomials bounded on the primes possible?

If $\{p_i\}$ is the sequence of all primes, is it possible that there exist a non constant $P\in \mathbb{Z}[x_1,\dots x_n]$ such that $P(p_i,p_{i+1},\dots p_{i+n-1})$ is bounded in $i$? More precisely …
Yaakov Baruch's user avatar
1 vote

polynomials with similar maxima-minima

I think that some computing around shows that $(x^6+y^6)^5$ and $(x^{10}+y^{10})^3$, or $x^{12}+y^{12}$ and $(x^4+y^4)^3$ for that matter, have the same set of extremes. So there can't be an algebraic …
Yaakov Baruch's user avatar
10 votes
1 answer
1k views

Is every positive polynomial the ratio of 2 positive coefficient polynomials?

I can prove that a quadratic positive polynomial is the ratio of 2 polynomials with non negative coefficients, for example $\displaystyle x^2-x+1/3=\frac{x^6+1/27}{x^4+x^3+2/3 x^2+1/3 x+1/9}$, and similarly … The full proof is not hard and involves some recursive polynomials related to Chebyshev polynomials of the second kind. …
Yaakov Baruch's user avatar
34 votes
1 answer
1k views

Does any cubic polynomial become reducible through composition with some quadratic?

irreducible cubic polynomial $P(X)\in \mathbb{Z}[X]$ is there always a quadratic $Q(X)\in \mathbb{Z}[X]$ such that $P(Q)$ is reducible (as a polynomial, and then necessarily the product of 2 irreducible cubic polynomials
Yaakov Baruch's user avatar
1 vote

Sum of two squares and implication of Bunyakovsky conjecture

Notice that in your example $(a1-\textit{i}*a2)*(-4+3\textit{i})/5 = a3-\textit{i}*a4$. $\mathbb{Z}[\textit{i}]$ is UFD and so is $\mathbb{Z}[\textit{i}][x]$.
Yaakov Baruch's user avatar
3 votes

q-th powers and roots of polynomials

A family of counterexamples is defined as follows: $r=2$, $p_1\ge 2, \quad p_2\ge 3, \quad d_1 \perp p_1, \quad d_2 \perp p_2$, $a=\dfrac{\tan(\pi d_1/p_1)-\tan(\pi d_2/p_2)}{\tan(\pi d_1/p_1)+\t …
Yaakov Baruch's user avatar
4 votes

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

Then naturally mathematicians became curious about whether the resulting enlarged family of numbers (or "field" in modern language) would A) consist always of solutions of polynomials; B) contain all such …
Yaakov Baruch's user avatar