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Just building on a previous question, where we showed that an odd number $p$ is prime in $\mathbb{Z}[i]$ if and only if $x^2+1$ does not have roots in $\mathbb{Z}/p\mathbb{Z}$.

The last part of this exercise is to show, in this case, that the multiplicative group of units of $\mathbb{Z}/p\mathbb{Z}$, denoted by $(\mathbb{Z}/p\mathbb{Z})^\times$ does not have an element of order 4, and $p \equiv 3 \mod 4$.

We can also conclude that $p \not\equiv 0,2 \mod 4$ because $p$ is odd.

Here is where I am getting a bit stuck. I assume the point is to show that if we assume that $p \equiv 1 \mod 4$, for example, $p = 5$, then $x^2+1$ has roots in $\mathbb{Z}/5\mathbb{Z}$. Here, $x^2+1 = 0 \mod 5$ for $x=2$. By the previous proposition, then $p$ is not prime in $\mathbb{Z}[i]$.

I am lacking the general statement of this proof. How can I go from assuming $p \equiv 1 \mod 4$ to $x^2+1 \equiv 0 \mod p$, in a general way?

jeffery_the_wind
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    $(\mathbf{Z}/p\mathbf{Z})^\times$ does not have order $p$, but rather $p-1$. – rae306 May 08 '20 at 13:50
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    $(\mathbb{Z}/p\mathbb{Z})^\times$ has order $p-1$ since it does not contain zero. – Adam Higgins May 08 '20 at 13:50
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    The thing is simply that a $x\in \mathbf{F}_p$ that satisfies $x^2=-1$ has order 4, so we must have $4\mid p-1 \iff p\equiv 1 \bmod{4}$. – rae306 May 08 '20 at 13:51
  • Oh my, OK thank you that was a bad mistake, I will work on correcting it. – jeffery_the_wind May 08 '20 at 13:51
  • @rae306 thank you that is very clear now, however now taking out my incorrect claim, it still seems we have not shown that there cannot be an element of order 4. – jeffery_the_wind May 08 '20 at 13:58
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    I don't understand what you mean. I have shown that if we have an element of order 4, then $p\equiv 1\bmod{4}$. But $p\equiv 3\bmod{4}$, so there can't be an element of order 4. – rae306 May 08 '20 at 14:02
  • @rae306 OK yes now i follow better your original comment. But then we still haven't shown that $p$ definitely needs to be congruent to $3 \mod 4$. Your just saying because it is not $3 \mod 4$, so it can't be true. Of course I understand this is a famously accepted result, I'm just trying to convince myself here. i guess it must be because $p \equiv 1 \mod 4$ implies there is a root $\mod p$. – jeffery_the_wind May 08 '20 at 14:06
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    $p$ is prime, so $p\not\equiv 0,2\bmod{4}$. Now, $\mathbf{F}_p^\times$ is cyclic of order $p-1$. So if $4\mid p-1$, then there exists an element $x$ of order $4$. This element satisfies $x^2=1$. If $p\equiv 3\bmod{4}$ then this has no solutions. – rae306 May 08 '20 at 14:09

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