# $\{r_1,r_2,…,r_{\phi(m)}\}$ is a reduced residue system modulo $m$ iff $\{r_1+k,r_2+k,…,r_{\phi(m)}+k\}$ be a reduced residue system modulo $m$

Let $2\lt m\in \mathbb N$ and $\{r_1,r_2,...,r_{\phi(m)}\}$ be a reduced residue system modulo $m$. I want to find a necessary and sufficient condition for $k$ such that the set $\{r_1+k,r_2+k,...,r_{\phi(m)}+k\}$ be a reduced residue system modulo $m$ to.

I now that:

Theorem: if $c_1,c_2,\ldots,c_{\varphi(m)}$ be the reduced residue set modulo $m>2$, then $$c_1+c_2+\cdots+c_{\varphi(m)} \equiv 0 \pmod{m}$$ (see here)

We want to the set $\{r_1+k,r_2+k,...,r_{\phi(m)}+k\}$ be a reduced residue system modulo $m$. So, by above theorem: $$r_1+k+r_2+k+...+r_{\phi(m)}+k\equiv 0 \pmod{m}$$ Because of the set: $\{r_1,r_2,...,r_{\phi(m)}\}$ is a reduced residue system modulo $m$, $$r_1+r_2+...+r_{\phi(m)}\equiv 0 \pmod{m}$$ So, we should have: $k\phi(m) \equiv 0 \pmod{m}$. Thus, the necessary condition is that: $k\phi(m) \equiv 0 \pmod{m}$. Is this condition sufficient? if not, what is the sufficient condition?

The condition $k\phi(m) \equiv 0 \bmod{m}$ is not sufficient.

Take $m=6$ and $k=3$.

Then $k\phi(m) = 3\cdot 2 \equiv 0 \bmod{m}$ but $\{1,5\}$ is a reduced residue system modulo $6$ and $\{1+3,5+3\} \equiv \{4,2\}$ is not a reduced residue system modulo $6$.

Some other examples are $m=10, k=5$ and $m=12, k=3$.

More generally, if $m=2^e n$, with $n$ an odd, take $k=n$. Then $k\phi(m) =n 2^{e-1}\phi(n) \equiv 0 \bmod{m}$ because $\phi(n)$ is even. But $1+n$ is even and so is not a unit mod $m$.

## The general case

Since the map $x\mapsto x+k$ is injective when considered as a map on $\mathbb Z/m$, it will induce a permutation of $U(m)$ if it sends units to units. So, the real question is:

When does $x\mapsto x+k$ preserve units mod $m$ ?

I don't have a complete proof right now, the answer seems to be this:

$x\mapsto x+k$ preserves units mod $m$ iff $k$ is a multiple of $\operatorname{rad}(m)$.

Here is a proof of one direction.

Take $k=tr$, a multiple of $r=\operatorname{rad}(m)$. Let $u$ be a unit and consider $u+k=u+tr$. If a prime divisor of $m$ divided $u+tr$, then it would divide $u$, which it doesn't. So $u+k$ is a unit whenever $u$ is unit.

I'd love to see a proof of the other direction.

As a consequence, there are no $k\ne0$ if $m$ is squarefree, because then $\operatorname{rad}(m)=m$ and $k \equiv 0 \bmod m$.

• – lhf Nov 26 '15 at 23:03