There exist integers $i, j > 0$ such that $x^i = y^j$;

(iii) There exist $z \in \A^+$ and integers $p, q > 0$ such that $x = z^p$ and $y = z^q$.

With these two lemmas we prove the following result on word equations involving the morphism $F$. Note that this result is within the general setting considered in [@JoFlMaNo], however we give an explicit solution of cases that we need later.

\[lem:words\_equations\_F\] Let $z$ and $p$ be nonempty words from $\{a,b,\alpha,\beta\}^+$.

(i) If $zp = F(p)F(z)$, then there is $x \in \{a,b,\alpha,\beta\}^+$ such that

    $z = x\bigl(F(x)x\bigr)^i$    and    $p = \bigl(F(x)x\bigr)^jF(x)$     for some $i,j \in \N$.

(ii) If $zp = pF(z)$, then there is $x \in \{a,b,\alpha,\beta\}^+$ such that

    $z = \bigl(F(x)x\bigr)^i$   and   $p =\bigl(F(x)x\bigr)^jF(x)$     for some $i,j \in \N$.

We prove Item $(i)$ by induction on $|zp| \geq 2$. If $|z| = |p|$, then $z = F(p)$ and the statement is true for $x = z$ and $i = j = 0$.

Assume $|z| > |p|$ (the case of $|z| < |p|$ is analogous). There must be a nonempty word $q$ such that $z = F(p)q$ and this yields $qp = F(z) = pF(q)$. By there are 