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Let $U(m)$ be the group of units mod $m$.

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

It is clear that $k$ must not be a unit: otherwise, $x=-k$ is sent to $0 \notin U(m)$.

I don't have a complete proof right now, but 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
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  • Follow-up from http://math.stackexchange.com/questions/1535307/r-1-r-2-r-phim-is-a-reduced-residue-system-modulo-m-iff-r-1. – lhf Nov 26 '15 at 20:21

2 Answers2

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If prime $p | m $ but NOT $p | k,$ then $k \neq 0 \pmod p.$ Take any unit $u,$ which will also not be divisible by $p.$ There is some multiple $u + wk \equiv 0 \pmod p,$ so $u+wk$ is not a unit

Will Jagy
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Let $p$ be a prime divisor of $m.$ Then the congruence class of $p$ (mod $m$) is not a unit, and therefore neither are the congruence classes of $p+k, p+2k,\ldots$

If $k$ would not be a multiple of $p$ then these congruence classes would span all congruence classes mod $m,$ leaving no units at all.

Thus $k$ is a multiple of all the prime divisors of $m.$

Justpassingby
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