Solutions in commutative ring.

**featherbox** · April 25th 2010, 05:23 AM

Show that the equation $x^{129}+x+2=0$ has precisely three solutions in the commutative ring $Z_{256}$ .

**tonio** · April 25th 2010, 08:49 AM

Originally Posted by featherbox

Show that the equation $x^{129}+x+2=0$ has precisely three solutions in the commutative ring $Z_{256}$ .

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

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

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

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

then $x=254$ .

Tonio

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