Peter Mueller
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 Feb 22 answered Does any polynomial with integer coefficients split over some prime field? Dec 13 revised How i show this beautiful inequality :$\frac{x^n}{x^m+y^m}+\frac{y^n}{y^m+z^m}+\frac{z^n}{z^m+x^m}\geq \frac{3} {2}(\frac{1}{\sqrt{3}})^{n-m}$? added 555 characters in body Dec 12 answered How i show this beautiful inequality :$\frac{x^n}{x^m+y^m}+\frac{y^n}{y^m+z^m}+\frac{z^n}{z^m+x^m}\geq \frac{3} {2}(\frac{1}{\sqrt{3}})^{n-m}$? Dec 9 comment Prime Divisors of an Integer Polynomial See Theorem 3 in projecteuclid.org/euclid.facm/1229442627. In even more down to earth terms: Let $\alpha$ and $\beta$ be roots of $f$ and $g$, respectively, $\mathbb Q(\gamma)=\mathbb Q(\alpha,\beta)$, and $h$ be the minimal polynomial of $\gamma$. There are polynomials $A$ and $B$ with $\alpha=A(\gamma)$, $\beta=B(\gamma)$. As $h$ is irreducible and $h(\gamma)=0=f(A(\gamma))$, we see that $h(X)$ divides $f(A(x))$ and $g(B(x))$. As $h(\mathbb Z)$ has infinitely many prime divisors, the assertion follows. Dec 9 comment Prime Divisors of an Integer Polynomial There is an elementary argument not using algebraic number theory as follows: If $f,g\in\mathbb Z[X]$ are non-constant polynomials, then there are infinitely many primes $p$ which divide some $f(a)$ and $g(b)$ for $a,b\in\mathbb Z$. Combine this with the other known elementary fact that the prime divisors of $\Phi_n(a)$ ($\Phi_n$ cyclotomic polynomial) divide $n$ or are $\equiv1\pmod{n}$. Mar 2 awarded Revival Dec 19 awarded Constituent Dec 15 awarded Caucus Oct 25 comment Centralizer of element in group PSL(2,F_p) abx identifies $PGL(2,\mathbb F_p)$ with the group of linear fractional maps from $\mathbb F_p\cup\{\infty\}$ to itself. Sep 24 awarded Autobiographer Aug 23 comment A Gap code for the alternating group $A_4$ If $G=AB$, then $G=ABg$ for any $g\in G$. So $B$ may be replaced with $Bg$. Now pick $g=b^{-1}$ with $b\in B$. Aug 22 awarded Critic Aug 22 comment A Gap code for the alternating group $A_4$ @Alexander Konovalov: I believe that you misunderstood the question. It is about $G=AB$, where $AB$ is the set of products $ab$ with $a\in A$, $b\in B$. So also your comment to my question doesn't make sense either. Aug 17 awarded Yearling Aug 15 answered A Gap code for the alternating group $A_4$ Feb 22 answered Normal subgroups of finite solvable groups Jun 14 comment Bunyakovsky conjecture I believe there is something wrong in Hagen's answer: Pick $f(X)=X(X-3)+1$. Then $\lvert f(3)\rvert=1=k$, but $3>2=\lvert f(0)\rvert+k$. Jun 14 answered Bunyakovsky conjecture Jun 14 awarded Revival Jun 14 awarded Supporter