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I'm reading "Elementary Number Theory by Kenneth H.Rosen", and in the book, there is one problem that I couldn't understand the solution.

Problem 9 (Chapter 9.2 - 6th edition)

Problem
Show that if $p$ is a prime and $p \equiv 1 \pmod{4}$, then there is an integer $x$ such that $x^2 \equiv -1 \pmod{p}$.

Solution from Textbook
By Larange's theorem, there are at most two solutions $x^2 \equiv 1 \pmod{p}$, and we know $x \equiv \pm1$ are the two solution.
Since $p \equiv 1 \pmod{4} \implies 4 | (p - 1) = \phi(p)$ so there is an element $x$ of order $4$ modulo p.
Then $x^4 = (x^2)^2 \equiv 1 \pmod{p}$, so $x^2 \equiv \pm1 \pmod{p}$. If $x^2 \equiv 1 \pmod{p}$ then $x$ does not have order $4$. Therefore $x^2 \equiv -1 \pmod{p}$.

What I did not understand is, "If $x^2 \equiv 1 \pmod{p}$ then $x$ does not have order $4$." I really don't see how the author came up with this one? Any idea?

Thanks,

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    if x^2 = 1 then x has order 2. (unless x = 1 then it has order 1)2011-04-01
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    @quanta: Many thanks! I must be blind :(.2011-04-01
  • 0
    For group-theoretic views of this see http://math.stackexchange.com/q/4872/2422011-04-01
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    @quanta Please consider converting your comment into an answer, so that this question gets removed from the [unanswered tab](http://meta.math.stackexchange.com/q/3138). If you do so, it is helpful to post it to [this chat room](http://chat.stackexchange.com/rooms/9141) to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see [here](http://meta.stackexchange.com/q/143113), [here](http://meta.math.stackexchange.com/q/1148) or [here](http://meta.math.stackexchange.com/a/9868).2013-06-15

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