bolton8l

2022-10-08

How to find the roots of $f(x)={x}^{2}+2x+2$ in ${\mathbb{Z}}_{3}$ ?

smh3402en

Beginner2022-10-09Added 11 answers

You can (in each case) start by realizing that

${x}^{2}+2x+2=0$

if and only if

$(x+1{)}^{2}=-1$

in whatever field you're working in. You've already noted that this doesn't work in

In ${\mathbb{Z}}_{3}$, this becomes

$(x+1{)}^{2}=2,$

but the only squares in ${\mathbb{Z}}_{3}$ are 0 and 1, so this is impossible. In ${\mathbb{Z}}_{5}$ becomes

$(x+1{)}^{2}=4.$

Now, ${2}^{2}={3}^{2}=4$, so 1 and 2 are the zeros of f(x) in ${\mathbb{Z}}_{5}$

${x}^{2}+2x+2=0$

if and only if

$(x+1{)}^{2}=-1$

in whatever field you're working in. You've already noted that this doesn't work in

In ${\mathbb{Z}}_{3}$, this becomes

$(x+1{)}^{2}=2,$

but the only squares in ${\mathbb{Z}}_{3}$ are 0 and 1, so this is impossible. In ${\mathbb{Z}}_{5}$ becomes

$(x+1{)}^{2}=4.$

Now, ${2}^{2}={3}^{2}=4$, so 1 and 2 are the zeros of f(x) in ${\mathbb{Z}}_{5}$

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