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I have a homework problem that I've attempted for days in vain... It's asking me to find an $n$ so that there is exactly one element of the complete residue system $\pmod n$ that is its own inverse apart from $1$ and $n-1$. It also asks me to construct an infinite sequence of $n's$ so that the complete residue system $\pmod n$ has elements that are their own inverses apart from $1$ and $n-1$.

For the first part, I tried all $n$ from $3$ up to $40$, but none worked... For the second part, I'm really confused...

Could someone please help me with this? Thanks!

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For the second: What about $3$ and $5 \pmod8$?

For the first: if $x^2\equiv 1$ then $(-x)^2\equiv 1$, too, so if there is only one (besides $\pm 1$), then $x\equiv -x \pmod{n}$, that is $2x=n$.

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  • $\begingroup$ If $n=2x$ then $x^2 \equiv 0 \text{ or } x \mod n$ $(=2x)$ $\endgroup$ – Mark Bennet Oct 28 '12 at 20:11
  • $\begingroup$ It shows that there is no such $n$. $\endgroup$ – Berci Oct 28 '12 at 20:20
  • $\begingroup$ This is really helpful, thanks very much! $\endgroup$ – hello.world Oct 28 '12 at 22:35
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Hint for the second part. You can choose the values of $n$ as you please and want to satisfy $x^2=kn+1$ for some integer $k$. Try simple cases first.

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