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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 :$\frac{x^n}{x^m+y^m}+\frac{y^n}{y^m+z^m}+\frac{z^n}{z^m+x^m}\geq \frac{3} {2}(\frac{1}{\sqrt{3}})^{n-m}$?
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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
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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