Find all the natural numbers N such that the reduced residue system consisting of least positive residues modulo N form an arithmetic progression.

Question

Firstly, I suppose that N is odd, then obviously the $1st$ term of the AP is $1$ and the $2nd$ term is $2$. Thus difference between $1st$ and $2nd$ term is $1$, thus the AP formed is $1, 2, 3, ... , N-1$. Thus $N$ must be prime. So we get that if $N$ is odd, then the R.R.S of $N$ forms an AP iff $N$ is prime. Now, suppose $N$ is even and not divisible by $3$, then $1st$ term is again $1$, $2nd$ term will be $3$. Here the common difference is $2$. Thus the AP formed is $1, 3, 5, ... , N-1$. This shows that $N$ must be a power of $2$. But, what if $N$ is even and divisible by $3$, or is there any better way to solve this?

Answer 1

Let $a_1, a_2,...., a_n$ be the reduced residue system modulo N in increasing order . All.primes and powers of 2 work so we assume N is non or those.then $a_n=N-1$ and $a_1=1$ , then common difference is $\frac{N-2}{n-1}=d$ . If N is even , then clearly so $(N, d)|(N, N-2)=2$ But that would imply there exists $x \in \mathbb{Z}$ such that $dx+1$[our terms in progression] ,such that if N even has a single odd factor then $dx+1$ is divisible by that factor[as d and that odd factor are coprime] making a contradiction, so N should be power of 2 .

if N is odd, then $a_2=2$ , hence all positive integers less than n are corpime to n, not possible as then N is a prime