# Has anyone studied this operator?

I've been studying a particular unary operator on the commutative ring $\mathbb{Z}/n\mathbb{Z}$. The operator is:

$\downarrow(x) = y\pmod{n}$ iff $n \equiv y \pmod{x}$, where $0< x,y \le n$.

The operator isn't distributive on addition or multiplication, so its use is probably fairly limited. I've proven a couple fairly trivial results, but I'm interested in seeing if there are more resources available.

One of the fairly trivial results: If $n$ is composite and $n=pq$ and WLOG $p \le q$, then if $q\ge kp+k$ I can show that

$(\downarrow(q-k))\mid p$.

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This is not really an operator on $\mathbb{Z}/n\mathbb{Z}$, it's an operator on the numbers $\{1,2,\ldots, n\}$. The reason is that if $x\equiv x'\pmod{n}$, we would expect any such operator to satisfy $\downarrow (x)=\downarrow(x')$, but this one doesn't.

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The definition of the operator can be extended trivially to satisfy this property, though. –  Foo Barrigno May 30 '13 at 18:59
Alas, it cannot. $n\equiv y\pmod{x}$ means $x|(n-y)$. $n\equiv y\pmod{x+n}$ means $(x+n)|(n-y)$. These two are rarely both true, and that's just for one choice of $x'$. Try some examples and you'll see. –  vadim123 May 30 '13 at 19:09
The "trivially" part was mapping x to the value I the range. In that sense, yes, it's an operator on the numbers {1,2,...,n}, but as long as you're rigorous about how the operator acts on each congruence class, it should be fine. –  Foo Barrigno May 30 '13 at 22:07
Yes. If the modulus is prime $\, n = p,\,$ then iterating the map $\ x\mapsto p\ {\rm mod}\ x\$ yields Gauss's algorithm for computing inverses mod $\,p.\,$ By this variant of the Euclidean algorithm, Gauss deduced the prime divisor property, i.e. if a prime divides a product then it divides some factor, which immediately yields uniqueness of prime factorizations (fundamental theorem of arithmetic).