LOL this language is messed up

Consider the symmetric group $S_4$. Choose $a=(1,2,3)$ and $b=(2,3,4)$. It is easy to check that $$a^3=\text{id}$$$$b^3=\text{id}$$$$ab=(1,2)(3,4)=b^2a^2$$$$ba=(1,3)(2,4)=a^2b^2$$So this means that the elements $a$ and $b$ satisfy the condition in the question. And since $ab\neq ba$, this means that the words $AB$ and $BA$ have different meanings. Remark: The idea to this problem is very similar to ISL 2005 C5

This problem is in Arthur Engel's book just with A and B replaced by 0 and 1 .

For those who are interested in "solving combinatorics by group theory", here are two more examples: 2018 Balkan Shortlist C3, smurfcc.

Can someone post a solution with an invariant formulation (I don't understand group theory)?

Bump. Invariant solution?

masked up group theory solution consider the following operations in the set $\{1,2,3,4\}$: $A: 1\rightarrow 2, 2\rightarrow 3, 3\rightarrow 1, 4\rightarrow 4$ $B: 1\rightarrow 1, 2\rightarrow 3, 3\rightarrow 4, 4\rightarrow 2$ you can easily check that starting from any number $x$: $\cdot A(A(A(x))=x$ $\cdot B(B(B(x))=x$ $\cdot A(A(B(B(x)))=A(B(x))$ then, the operations satisfy the condition in the question, but: $3=A(B(1))\neq B(A(1))=2$

Yup it's just masked up A BRILLIANT solution indeed! small question: how do you come up with a solution like that in a test?

a22886 wrote: small question: how do you come up with a solution like that in a test? As I said it's just a masked up group theory solution, so you'll have to see something like this before the test. But just try seeking operations that satisfy the question and you will eventually find it