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Assume that $p \equiv 1 \pmod4$, where $p$ is prime. The multiplicative group of the finite field $\mathbb{Z_p}$ is cyclic and is of order $p-1$. We know that $4$ is a divisor of $p-1$.

Now my text claims that $\mathbb{Z_p}$ contains an element $n$ of multiplicative order $4$. Why is that?

If I have understood it correctly it means that we have $n^4 \equiv 1 \pmod p$ for some $n \in \mathbb{Z_p} $ Why?

Also, why does it follow that $n^2$ has multiplicative order $2$?

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    $\begingroup$ A small nitpick: $n^4=1$ for some $n\in {\bf Z}_p$. This is equivalent to saying that $n^4\equiv 1 (\mod p)$ for some $n\in {\bf Z}$. The multiplication in ${\bf Z}_p$ is already modulo $p$. What you've written is not exactly nonsense (you could interpret it as congruence modulo the zero ideal $(p)=(0)$, which is just trivial), but pretty close. $\endgroup$
    – tomasz
    Dec 13 '13 at 19:46
  • $\begingroup$ @tomasz Thank you for clearing that up. so I just just say "..for some positive integer n" rather than $n$ as a element in the group? $\endgroup$
    – John Smith
    Dec 13 '13 at 20:12
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    $\begingroup$ Yes, that's correct. If you're familiar with quotients and ideals, you can say that $(r+I)^4=1+I$ in $R/I$, or equivalently that $r^4\equiv 1(\operatorname{mod} I)$ which is just the same as $r^4\in 1+I$. But $(r+I)^4\equiv 1+I(\operatorname{mod} I)$ is not a very good notation. We think of elements of ${\bf Z}_p$ as elements of ${\bf Z}$, but that's not really true. $\endgroup$
    – tomasz
    Dec 13 '13 at 21:42
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The multiplicative group is cyclic and of order $p - 1$. Let $g$ be a generator; and look at the element a quarter way down the list $1, g, g^2, \dots, g^{p-2}$ of all elements, i.e, look at $h := g^{(p-1)/4}$. That element has order 4, as can be seen by explicit computation.

Note, by the way, that $h^2 = g^{(p-1)/2}$ is order 2 and therefore must be (the residue class of) $-1$.

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