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Acer A200 Simple Root V3 96 ⚓





 
 
 
 
 
 
 

Acer A200 Simple Root V3 96

I was looking for a new method to disable the booting from the acer recovery so that i could safely root my acer before .Q:

Classifying $\mathbb{Z}_5[i]$ over $\mathbb{Z}_5$

I think the minimal polynomial of the complex number $\sqrt[5]{3}+i\sqrt[5]{2}$ is $x^5-3x^3+2x$ over $\mathbb{Z}_5$.
I want to understand the following, is it correct that I can express $\sqrt[5]{3}+i\sqrt[5]{2}$ over $\mathbb{Z}_5$ as a linear combination of $1,i,i^2$?
The minimal polynomial over $\mathbb{Z}_5$ is the smallest polynomial $g$ such that $g(\sqrt[5]{3}+i\sqrt[5]{2})=0$, and the minimal polynomial of $\sqrt[5]{3}+i\sqrt[5]{2}$ over $\mathbb{Z}_5$ is $x^5-3x^3+2x$, hence $\sqrt[5]{3}+i\sqrt[5]{2}$ is algebraic over $\mathbb{Z}_5$ and can be expressed as a linear combination of $1,i,i^2$
Thanks.

A:

The minimal polynomial over $\mathbb Z_5$ must be of the form $x^5-ax^4+bx^3-bx^2+cx-c$ where $a,b,c \in \mathbb Z_5$. The minimal polynomial of $\sqrt[5]{3}+i\sqrt[5]{2}$ over $\mathbb Z_5$ is a polynomial of the form $x^5-3x^3+2x=x^5+x^3+1$. So you want to find $a,b,c\in \mathbb Z_5$ such that $a+b+1=3,\;a+c=2,\;b

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