Roots of $x^2-x+2=0 \in \mathbb{Z}_3[i]$

Question

I've been challenged by a professor to find the roots of $x^2-x+2=0$ in the field $\mathbb{Z}_3[i] = \{a+bi \; \vert \; a,b \in \mathbb{Z}_3\}$. I used the "normal" quadratic formula and got roots of $2+2i$ and $2+i$, and was told there's a way without using the formula. Is there a method for this besides guessing and checking the nine elements of $\mathbb{Z}_3[i]$?

Answer 1

Completing the square (which leads to the quadratic formula):

$x^2-x+2=0$ $\iff x^2+2x+2=0$ $\iff (x+1)^2=-1$ $\iff x+1=\pm i$

$\iff x=2\pm i$ $\iff x=2+i\text{ or }x=2+2i,$ since $-1=2 $ in $\mathbb Z_3$.

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Alternatively, note that $0, 1, $ and $2$ are not roots, so the roots are complex conjugates $a\pm bi$.

The sum of the roots of $x^2-tx+d$ is $t$, so we have $(a+bi)+(a-bi)=2a=1;$ i.e., $a=2$.

The product of the roots of $x^2-tx+d$ is $d$, so we have $(a+bi)(a-bi)=a^2+b^2=2; $

i.e., $b^2=1$. Again we get the roots $a\pm bi=2\pm i$.

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Really, though, checking the nine elements $\{0,\pm1,\pm i, \pm 1\pm i\}$ of $\mathbb Z_3[i]$ is not hard.

It is rather easy to see that $0$ and $\pm 1$ are not roots.

Nor is $\pm i$ since $i^2\mp i+2$ has an imaginary part.

Nor is $1+i$, since $(1+i)^2-(1+i)+2=2i-1-i+2$ also has an imaginary part,

so the roots must be the other pair of conjugates $-1\pm i$.

Answer 2

In general it's better to think of $2=-1$. That way multiplication and exponentiation became a matter of just units, negatives, and zeros and magnitudes and growing values are just taken out of the picture altogether. Also noting that $1$ and $-1$ are "natural opposites" is just easy to intuit than $1$ and $2$.

So that said.....

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Okay. This may be kind of dumb but:

$x^2 -x +2 = 0\iff x^2-1 = x$

And

$x^2 -1=x$

$(x^2 -1)^2 = x^2$

$x^4 -2x^2 +1 = x^2$

$x^4 = -1$

$x^2 =\pm i$

And so $x^2 -1 = x\implies \pm i - 1 =x$

.... and that's it. $x = -1 \pm i$.

However when we squared both sides we potentially added extraneous solutions.

So we must test these which easy enough.

(I had originally wanted to comment about roots coming in conjugate pairs but now that I've solved quicker than I expected we don't really need to.)

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Not sure that a better way but I can say it didn't use the quadratic formula.

I wouldn't try it with any field more complex than $\mathbb Z_3[i]$. The "all or nothing" nature of $Z_3$ allowed us to make a lot of draconian "either 0 or unit; either positive or negative" absolutes which wouldn't really occur in more "field-like" fields.