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So obviously $6$ isn't a quadratic residue $mod(7)$ thus there are no zeros in $\Bbb Z_7$. So what I did next is considered considered the field $\Bbb Z_7[x]/<x^2+1>$ obviously $x+<x^2+1>$ is a zero for the polynomial so we will denote that by $i$, keep in mind that $i^2=6$. My question is really about the field $\Bbb Z_7(i)$, that is the smallest field containing all of $\Bbb Z_7$ and $i$. Well since $i^2=6$ this field has all its elements of the form $a+bi$ where $a,b \in \Bbb Z_7$ so that field contains only 14 elements which isn't a power of a prime. I must be doing something wrong since I am a bit rusty on algebra, can someone please point out my mistake.

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    $\begingroup$ Your field contains $49$ elements, as you can choose $7$ for $a$ and $b$. $\endgroup$ – Krijn Sep 20 '15 at 14:44
  • $\begingroup$ Right of course, I'm an idiot as usual. Thanks for the answer $\endgroup$ – H_Hassan Sep 20 '15 at 14:45
  • $\begingroup$ Relax, every one of us has made foolish errors (and most of us continue to do so). The point is to learn from our errors, and not make the same mistake twice. $\endgroup$ – Lubin Sep 21 '15 at 3:13
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$(a+bi)^2 = a^2 - b^2 + 2abi = -1$ only iff $2ab = 0$, which gives us that either $a = 0$ or $b=0$. You have excluded the options where $b=0$, which leaves us with $a = 0$. Here we get the solutions $b = 6$ and $b = 1$.

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