2015-12-4

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Problema:} Fie Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle K} un corp cu Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m \geq 2} elemente si Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f \in K[X]} . Aratati ca urmatoarele afirmatii sunt echivalente:

$(i)$ Exista $g\in K[X]$ astfel incat $f(X)=g(X^{m-1})$;

$(ii)$ Pentru orice $a\in K^{*}$ avem $f(X)=f(aX)$.

$Solutie$

$(i)\rightarrow (ii)$ Din teorema lui Lagrange aplicata grupului $(K^{*},\cdot )$ avem ca $x^{m-1}=1,\forall x\in K^{*}$, deci $f(aX)=g((aX)^{m-1})=g(a^{m-1}X^{m-1})=g(X^{m-1})=f(X)$.

$(ii)\rightarrow (i)$ $(Robert\ Rogozsan)$

Ne folosim de urmatoarea

$Lema:$ Fie Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F} un corp finit cu Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n} elemente. Atunci Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{a \in F}a^k= \begin{cases} 0 & \text{, dacă } n \text{divide pe} k f(c) & \text{, dacă } n \text{nu divide pe} k \end{cases}}