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\begin{document}

\section{Equation 1526}

I work with the free multiplicative group $G$ on countably many generators as usual.  The functional equation is
\[
f(f(f(h)h^{-1})f(1)^{-1})=h^{-1}f(1)^{-1}.
\]
Think of this as saying that if $(1,x_1),(a,b),(ba^{-1},c)\in f$, then we must have $(cx_1^{-1},a^{-1}x_1^{-1})\in f$.

Define $\mathscr{E}$ as the collection of sets $E\subseteq G^2$ satisfying the following properties.
\begin{itemize}
    \item[(1)] $E$ is finite.
    \item[(2)] $E$ is a function.
    \item[(3)] $(1,x_1),(x_1,x_2),(x_2x_1^{-1},x_1^{-1}),(x_1^{-2},x_1^{-2})\in E$ (where $x_1,x_2$ are distinct generator).
    \item[(4)] If $(a,b),(ba^{-1},c)\in E$, then $(cx_1^{-1},a^{-1}x_1^{-1})\in E$.
    \item[(5)] If $(a,b),(c,d)\in E$ and $ab^{-1}=cd^{-1}$, then $(a,b)=(c,d)$.
\end{itemize}

\begin{lemma}
For any $E\in \mathscr{E}$ and any $a\in G$, there is an extension $E\subseteq E'\in \mathscr{E}$ where the functional equation holds for $a$.
\end{lemma}
\begin{proof}
{\bf Case 1}: Assume $(a,b)\in E$ for some $b\in G$.

If $ba^{-1}\in {\rm dom}(E)$, then by condition (4) we are already done.  So reduce to the case when $ba^{-1}\notin {\rm dom}(E)$.  Let $c$ be any generator not appearing anywhere in $E$, and fix
\[
E':=E\cup\{(ba^{-1},c),(cx_1^{-1},a^{-1})x_1^{-1}\}.
\]
Conditions (1), (2), and (3) on $E'$ are clear, as is condition (5) from the newness of $c$.  

For condition (4), we need to note that $a\neq 1$ since $ba^{-1}\notin {\rm dom}(E)$.  Thus, from (5) for $E$, we have $ba^{-1}\neq x_1$.  A case analysis now suffices (using (5) to show that we don't cause any problems with the old pairs in $E$).

{\bf Case 2}: Assume $a\notin {\rm dom}(E)$.  If there is some $(x,y)\in E$ with $a=yx^{-1}$, then use Case 1 to make the functional equation hold for $x$, and then $a$ will belong to the domain of that extension, and we reduce to Case 1 again.

So, we may assume there is not such pair $(x,y)$.  Taking $b$ to be any generator of $G$ not appearing in $E$ or in $a$, then $E\cup\{(a,b)\}\in \mathscr{E}$ and we again revert to Case 1.
\end{proof}

With the choice that $f(1)=x_1$, the functional equation for 1223 is 
\[
f(f(x_1^{-1}f(x_1^{-1})^{-1})f(x_1^{-1})x_1)=1.
\]
The initial seed
\[
\{(1,x_1),(x_1,x_2),(x_2x_1^{-1},x_1^{-1}),(x_1^{-2},x_1^{-2}),(x_1^{-1},x_3),(x_1^{-1}x_3^{-1},x_4),(x_4x_3x_1,x_5),(x_5x_1^{-1},x_3)\}
\]
works to contradict this equation.

The functional equation for 2744 is
\[
f(x_1^{-1}f(f(h)x_1^{-1})^{-1})=hx_1^{-1}f(f(h)x_1^{-1})^{-1}.
\]
The seed
\[
\{(1,x_1),(x_1,x_2),(x_2x_1^{-1},x_1^{-1}),(x_1^{-2},x_1^{-2}),(x_3,x_4),(x_4x_1^{-1},x_5),(x_1^{-1}x_5^{-1},x_6)\}
\]
works.
\end{document} 

