If $v^6w^8 = w^{12}v^4$ then $(vw)^2 = v^2w^2$?

Question

Given the words $v,w \in \sum^*$, is this correct?

1. If $v^6w^8 = w^{12}v^4$ then $(vw)^2 = v^2w^2$

2. If $vw^2 = wv^2$ then $v=w$

For one, I tried $v=\epsilon, w=\epsilon$ and it worked, and I also tried $v=aa, w= aaaa$ and it also worked, but I am not quite sure how to prove it.

Regarding two, I could not come up with examples except $v,w=\epsilon$.

Any help is appreciated.

Answer 1

In case $1$, taking lengths you have

$$ 6{\sf length}(v)+8{\sf length}(w)=12{\sf length}(w)+4{\sf length}(v) $$

so ${\sf length}(v)=2{\sf length}(w)$. So looking at the $12{\sf length}(w)$ leftmost charcaters in the word $v^6w^8=w^{12}v^4$, we see that $v^6=w^{12}$, so $v=w^2$.

Similarly in case $2$, loking at lengths you see that $v$ and $w$ have the same length, and looking at the ${\sf length}(v)$ leftmost charcaters in the word $vw^2=wv^2$ you see that $v=w$.

Answer 2

The answer is yes in both cases. You could use Corollary 1.2.6 of Lothaire's book Combinatorics on Words, which says:

If you have a nontrivial relation between two words $v$ and $w$, then both words are powers of the same word $u$.

Thus $v = u^a$ and $w = v^b$ for some positive integers $a$ and $b$ and your questions become trivial.