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This question is a follow-up and analogue of a previous very similar question. I have to seperate questions since it is a bit of a rule ; 1 question at a time , hence this seperate one. Looking at the other analogue question is not neccessary and can be ignored but for those who care here it is :

About 2 operators and $A*(B+C) =(A+B)*(A+C)$

Anyway,

Consider $2$ distinct binary operators defined for a finite set with $n$ elements.

Operator $*$ behaves like a commutative latin quandle :

$$x*x = x$$

$$a*b=b*a$$

$$a*(b*c)=(a*b)*(a*c)$$

And forms a latin square.

Commutative operator $+$ behaves like

$$A*(B+C) =(A*B)+(A*C)$$

$$X+Y = Y+X$$

And forms a latin square.

Question :

Is $+$ neccessarily associative ?

What are examples of strict non-associative or strict associative ones if they exist ? ( strict mean always that property for every pair of elements )


The special cases where the quandle has no subquandle is perhaps interesting. ( this happens if $n$ is prime but that is not a neccessary condition , it is similar to subgroups basically )


ps: do not confuse with bi-quandles as most define them.


Remarks :

Also note the number of solutions $(*,+)$ for a set with $n$ elements is a bit of a mystery for me , Im not even sure with all these restrictions given that every $n$ has such a pair of operators. Not even for sufficiently large $n$.

But that is not the main question but just related. Unless there are no solutions $(*,+)$ for $n$ or only for finitely many $n$ then it becomes very important.


mick
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    You may have noticed this but: Consider $a(a+a)=aa+aa=a+a$. Since $$ is a Latin square, there is exactly one $x$ st $x(a+a)=a+a$; and we know that $(a+a)*(a+a)=a+a$; thus $x=a+a$ and so $a=a+a$. – Alex K Jan 30 '23 at 02:38
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    What if $+$ is the same as $*$? – Alex K Jan 30 '23 at 03:48
  • @AlexK yes good point. I forgot to mention distict. I thought I did , but I did not lol. Anyway if $+ = *$ that is not so interesting. It literally reduces to a single operator then. – mick Jan 31 '23 at 23:28

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