This post was inspired by an exchange with the indian woman mathematician Ajit Iqbal Singh.

The chinese remainder theorem can be stated as follows:

Let $n_1, \dots, n_r \ge 2$ be positive integers such that $n_i \wedge n_j = 1$ iff $i \neq j$, then the system of equations $$ x \equiv a_i [n_i] \text{, for } i=1, \dots , r$$ admits a solution $x$ which is unique modulo $\prod_i n_i$.

There is a natural way to generalize this statement to the cyclic subfactor planar algebras, as follows

(I warn the reader who don't know the subfactor planar algebra theory, that I've reformulated the statement in the case of finite index inclusions of groups, just after).

Let $P$ be a finite index irreducible subfactor planar algebra, and assume that $P$ is cyclic, i.e. admits a *distributive* biprojections lattice. The statement is the following:

Let $b_1, \dots , b_r \not \in \{ e_1, id \}$ be a subset of biprojections satisfying $ b_i \vee b_j = id$ iff $i \neq j$, then the system:
$$ b_i * x * b_i \sim b_i * a_i * b_i \text{, for } i=1, \dots ,r $$
(with $a_i, x$ positive operators) admits a solution $x$ which is unique (up to $\sim$) as $B * x * B$, with $B = \bigwedge_i b_i$.

Now in the group framework: let $(H \subset G)$ be a finite index inclusion of groups, such that the interval lattice $[H,G]$ is *distributive*, then the statement is the following:

Let $K_1, \dots , K_r$ be strict intermediate subgroups (i.e. $H \subsetneq K_i \subsetneq G$, $\forall i$), such that $\langle K_i , K_j \rangle = G$ iff $i \neq j$, then the system of double-coset equations:
$$ K_ixK_i = K_ig_iK_i \text{, for } i=1, \dots ,r $$
admits a solution $x$ which is unique as $KxK$ with $K = \bigcap_i K_i$.

*Remark*: This group statement above with $G = \mathbb{Z}$, $K_i = n_i\mathbb{Z}$ and $H = K = \prod_i n_i \mathbb{Z}$ is equivalent to the classical statement of the chinese remainder theorem.

**Question**: Is the chinese remainder statement true for the cyclic subfactor planar algebras?

*Remark*: A counter-example for an inclusion of finite groups gives a counter-example for the subfactors planar algebras. I will edit a ckeck by GAP soon.

*Remark*: The *distributivity* assumption is very useful for generalizing the beginning of the proof here.

*Motivation*: Try to better understand in what sense the cyclic subfactor theory is a *quantum arithmetic*.