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Let $n\ge 2$ be a natural number.

Then, every possible order of an number modulo $n$ is a divisor of $\lambda(n)$. $\lambda(n)$ is the Carmichael-function, it is the largest possible order modulo $n$.

Which numbers $n$ have the following property : For every $d|\lambda(n)$, except $d=1$, there is a number $a$ with $1<a<n$, such that the order of $a$ modulo $n$ is $d$ ?

Additionally : If we are looking for a number a with order $2$, for which $n$ can we avoid $a=n-1$, which is trivial ?

I experimented with PARI/GP and it seems that the desired property is fulfilled, if there is a number $a$ with order $\phi(n)$ modulo $n$.

Peter
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  • See here for some stuff about finding square roots of $1$ modulo $n$. Also here, here or here. This is one of the more frequently recurring themes. – Jyrki Lahtonen Nov 20 '15 at 10:37
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    But did I understand the main question correctly? If $a$ is of order $\lambda(n)$, and $d\mid\lambda(n)$, then $a^{\lambda(n)/d}$ is automatically of order $d$, right? – Jyrki Lahtonen Nov 20 '15 at 10:39
  • How can we be sure, that the order of $a^{\lambda(n)/d}$ is not smaller than $d$ ? – Peter Nov 20 '15 at 10:46
  • If that were the case, then the order of $a$ would also be less than $\lambda(n)$. – Jyrki Lahtonen Nov 20 '15 at 10:49
  • The general fact that holds in all groups. If $a$ is of order $m$, then $a^n$ is of order $m/\gcd(m,n)$. It suffices to prove this in a cyclic group, because everything takes place inside the cyclic subgroup generated by $a$. – Jyrki Lahtonen Nov 20 '15 at 10:49
  • Do you mean the following : We choose a number $a$ with order $\lambda(n)$ modulo $n$ and simply choose $a^{\lambda(n)/d}$ ? – Peter Nov 20 '15 at 10:50
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    Yes. ${}{}{}{}$ – Jyrki Lahtonen Nov 20 '15 at 10:51
  • @Jyrki Lahtonen. Yes! The integers from 1 to n-1 that are co-prime to n are a group under multiplication mod n. That answers the Q :Every n .For the second part it is required that $n$ is composite and that $nx+1$ is a square for at least one $x\in {1,...,n-1}$. E.g. n=21 and x=3. I see you have given 2 links about this. – DanielWainfleet Nov 20 '15 at 10:56
  • Seems convincing, it remains to solve the case $d=2$. When can we avoid $a=n-1$ ? – Peter Nov 20 '15 at 10:56
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    Peter, the congruence $x^2\equiv1\pmod n$ has $2^{t+e}$ pairwise incongruent solutions, where $t$ is the number of odd prime factors of $n$. The extra $e$ depends from the power of $2$ dividing $n$. If $n$ is not divisible by four, then $e=0$. If $n$ is divisible by four but not by eight, then $e=1$. If $8\mid n$, then $e=2$. That's more or less covered in the questions I linked. Chinese remainder theorem reduces the problem to the case where $n$ is a prime power, and this is what you get. You can avoid $a=n-1$, whenever $t+e>1$. – Jyrki Lahtonen Nov 20 '15 at 15:58
  • Thank you very much, Jyrki! – Peter Nov 20 '15 at 19:07
  • One more thing : Is there always a number $a$ with order $\lambda(n)$ modulo $n$ ? – Peter Nov 21 '15 at 09:07

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