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Thread: Solutions in commutative ring.

  1. #1
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    Solutions in commutative ring.

    Show that the equation $\displaystyle x^{129}+x+2=0$ has precisely three solutions in the commutative ring $\displaystyle Z_{256}$.
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  2. #2
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    Quote Originally Posted by featherbox View Post
    Show that the equation $\displaystyle x^{129}+x+2=0$ has precisely three solutions in the commutative ring $\displaystyle Z_{256}$.

    The multiplicative group of units of that ring is $\displaystyle U:=U(\mathbb{Z}_{256})$ whose order is $\displaystyle \phi(256=2^8)=2^7=128\Longrightarrow u^{128}=1\,\,\,\forall\,u\in U$

    $\displaystyle \Longrightarrow \,\forall x\in U\,,\,\,x^{129}+x+2=2x+2=2(x+1)=0\!\!\!\pmod {256}$ $\displaystyle \iff x=-1\!\!\!\pmod{256}$ or $\displaystyle x+1=128\!\!\!\pmod {256}\iff x=127,\,255\!\!\!\pmod{256}$ , and

    these are the only solutions with $\displaystyle x\in U$


    No, if $\displaystyle x=2k\notin U$ is a solution then $\displaystyle x^{129}+x+2=2^{129}k^{129}+2k+2=\left(2^8\right)^{ 16}\cdot 2k^{129}+2k+2$ $\displaystyle =2(k+1)=0\!\!\!\pmod {256}\iff k+1=128\iff k=127$ and

    then $\displaystyle x=254$.

    Tonio
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