Given is the number $n$ which is divisible by the square of a prime number $p$. Let a positive integer $a$ be euler-course when $n\mid (a^n -a)/a$. Show that $(1,2, ...,n)$ has at most $\varphi(n)/2$ of $\varphi(n)$-course numbers.
I was trying to solve this using primitive roots, by looking at the primitive root of $p^2$, but I couldn't quite do it.
Any hints or solutions are greatly appreciated.
Clearly, $n | (a^n-a)/a=a^{n-1}-a \implies ord(a) | n-1$.
Also, $ord(a) | \lambda(n)$ where $\lambda(n)$ is the Carmichael function.
So, $ord(a) | \gcd(\lambda(n),n-1)=d$ (say).
Now, since $n$ is not square-free, $\lambda(n) \not | n-1$, as proven here.
Hence, $d$ is a proper divisor of $\lambda(n)$, so $d < \lambda(n)$.
Now, $a$ is a root of $x^d-1$, and the roots of $x^d-1$ form a subgroup of $\mathbb Z_n ^ \times$. The subgroup is proper, since $d < \lambda(n)$, S (say). So, $|S| \leq \lambda(n)/2 \leq \phi(n)/2$ i.e. the maximum possible number of such $a$'s is $\phi(n)/2$.
