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\begin{document}

\section{Equation 917, Hardy}

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

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)] $(1,x_1)\in E$ (where $x_1$ is a generator).
    \item[(4)] If $(a,b),(ba^{-1}x_1^{-1},c)\in E$, then $(cx_1,a^{-1})\in E$.
    \item[(5)] If $(a,b)\in E$, then $ab^{-1}\neq x_1$.
    \item[(6)] 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 $ba^{-1}x_1^{-1}\in {\rm dom}(E)$, then by condition (4) we are already done.  So reduce to the case when $ba^{-1}x_1^{-1}\notin {\rm dom}(E)$.  Let $c$ be any generator not appearing anywhere in $E$, and fix
\[
E':=E\cup\{(ba^{-1}x_1^{-1},c),(cx_1,a^{-1})\}.
\]
Conditions (1) and (3) on $E'$ are clear.  Condition (2) on $E'$ follows from (6) for $E$.  Condition (4) on $E'$ comes from condition (5) on $E$.  Condition (5) is easy, and so is condition (6).

{\bf Case 2}: Assume $a\notin {\rm dom}(E)$.  If there is some $(x,y)\in E$ with $a=yx^{-1}x_1^{-1}$, 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 1629 is $f^2(x_1^{-1})=x_1^{-1}$.  The initial seed
\[
\{(1,x_1),(x_1^{2},1),(x_1^{-1},x_2),(x_2,x_3),(x_3x_1,x_1)\}\in \mathscr{E}
\]
works to contradict this equation.

The functional equation for 2441 is $f(f(f(x_1^{-1})x_1)^{-1})=f(f(x_1^{-1})x_1)^{-1}$.  The initial seed
\[
\{(1,x_1),(x_1^{2},1),(x_1^{-1},x_2),(x_2x_1,x_3),(x_3^{-1},x_4)\}
\]
works.
\end{document} 

