%root: main.tex
\subsection{Proof of \Cref{th:single-p-hard}}
\begin{proof}
For the sake of contradiction, assume that for any $G$, we can compute $\rpoly_{G}^3(\prob,\dots,\prob)$ in $o\inparen{m^{1+\eps_0}}$ time.
Let $G$ be the input graph. It is easy to see that one can compute the expression tree for $\poly_{G}^3(\vct{X})$ in $O(m)$ time. Then by \Cref{th:single-p} we can compute $\numocc{G}{\tri}$ in further time $o\inparen{m^{1+\eps_0}}+O(m)$. Thus, the overall, reduction takes $o\inparen{m^{1+\eps_0}}+O(m)= o\inparen{m^{1+\eps_0}}$ time, which violates \Cref{conj:graph}.
\qed
\end{proof}

\subsection{Tools to prove \Cref{lem:lin-sys}}

Note that $\rpoly_{G}^3(\prob,\ldots, \prob)$ as a polynomial in $\prob$ has degree at most six. Next, we figure out the exact coefficients since this would be useful in our arguments:
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\begin{Lemma}\label{lem:qE3-exp}
%When we expand $\poly_{G}^3(\vct{X})$ out and assign all exponents $e \geq 1$ a value of $1$, we have the following result,
For any $\prob$, we have:
	{\small 
	\begin{align}
		\rpoly_{G}^3(\prob,\ldots, \prob) &= \numocc{G}{\ed}\prob^2 + 6\numocc{G}{\twopath}\prob^3 + 6\numocc{G}{\twodis}\prob^4 + 6\numocc{G}{\tri}\prob^3\nonumber\\
		&+ 6\numocc{G}{\oneint}\prob^4 + 6\numocc{G}{\threepath}\prob^4 + 6\numocc{G}{\twopathdis}\prob^5 + 6\numocc{G}{\threedis}\prob^6.\label{claim:four-one}
	\end{align}}
\end{Lemma}
\subsubsection{Proof for \Cref{lem:qE3-exp}}
\begin{proof}%[Proof of \Cref{lem:qE3-exp}]
By definition we have that
		\[\poly_{G}^3(\vct{X}) = \sum_{\substack{(i_1, j_1), (i_2, j_2), (i_3, j_3) \in E}}~\; \prod_{\ell = 1}^{3}X_{i_\ell}X_{j_\ell}.\]
Hence $\rpoly_{G}^3(\vct{X})$ has degree six. Note that the monomial $\prod_{\ell = 1}^{3}X_{i_\ell}X_{j_\ell}$ will contribute to the coefficient of $\prob^\nu$ in $\rpoly_{G}^3(\vct{X})$, where $\nu$ is the number of distinct variables in the monomial.
		%Rather than list all the expressions in full detail, let us make some observations regarding the sum.  
Let $e_1 = (i_1, j_1), e_2 = (i_2, j_2), and e_3 = (i_3, j_3)$.  
We compute $\rpoly_{G}^3(\vct{X})$ by considering each of the three forms that the triple $(e_1, e_2, e_3)$ can take.

\textsc{case 1:} $e_1 = e_2 = e_3$ (all edges are the same).  When we have that $e_1 = e_2 = e_3$, then the monomial corresponds to $\numocc{G}{\ed}$. There are exactly $\numedge$ such triples, each with a $\prob^2$ factor in $\rpoly_{G}^3\left(\prob,\ldots, \prob\right)$.

\textsc{case 2:}  This case occurs when there are two distinct edges of the three, call them $e$ and $e'$.  When there are two distinct edges, there is then the occurence when $2$ variables in the triple $(e_1, e_2, e_3)$ are bound to $e$.  There are three combinations for this occurrence in $\poly_{G}^3(\vct{X})$.  Analogusly, there are three such occurrences in $\poly_{G}^3(\vct{X})$ when there is only one occurrence of $e$, i.e. $2$ of the variables in $(e_1, e_2, e_3)$ are $e'$.  %Again, there are three combinations for this.  
This implies that all $3 + 3 = 6$ combinations of two distinct edges $e$ and $e'$ contribute to the same monomial in $\rpoly_{G}^3$. % consist of the same monomial in $\rpoly$, i.e. $(e_1, e_1, e_2)$ is the same as $(e_2, e_1, e_2)$.  
Since $e\ne e'$, this case produces the following edge patterns: $\twopath, \twodis$, which contribute $6\prob^3$ and $6\prob^4$ respectively to  $\rpoly_{G}^3\left(\prob,\ldots, \prob\right)$.

\textsc{case 3:} All $e_1,e_2$ and $e_3$ are distinct.  For this case, we have $3! = 6$ permutations of $(e_1, e_2, e_3)$, each of which contribute to the same monomial in the \textsc{SMB} representation of $\poly_{G}^3(\vct{X})$.  This case consists of the following edge patterns: $\tri, \oneint, \threepath, \twopathdis, \threedis$, which contribute $6\prob^3, 6\prob^4, 6\prob^4, 6\prob^5$ and $6\prob^6$ respectively to  $\rpoly_{G}^3\left(\prob,\ldots, \prob\right)$.
\qed
\end{proof}

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Since $\prob$ is fixed, \Cref{lem:qE3-exp} gives us one linear equation in $\numocc{G}{\tri}$ and $\numocc{G}{\threedis}$ (we can handle the other counts due to equations (\ref{eq:1e})-(\ref{eq:3p-3tri})). However, we need to generate one more independent linear equation in these two variables. Towards this end we generate another graph related to $G$:
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Next, we relate the various sub-graph counts in $\graph{2}$ to $\graph{1}$ ($G$).

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\begin{Lemma}\label{lem:3m-G2}
The $3$-matchings in graph $\graph{2}$ satisfy the identity:
\begin{align*}
\numocc{\graph{2}}{\threedis} &= 8 \cdot \numocc{\graph{1}}{\threedis} + 6 \cdot \numocc{\graph{1}}{\twopathdis}\\
&+ 4 \cdot \numocc{\graph{1}}{\oneint} + 4 \cdot \numocc{\graph{1}}{\threepath} + 2 \cdot \numocc{\graph{1}}{\tri}.
\end{align*}
\end{Lemma}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{Lemma}\label{lem:tri}
For $\ell > 1$ and any graph $\graph{\ell}$, $\numocc{\graph{\ell}}{\tri} = 0$.
\end{Lemma}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Proof of \Cref{th:single-p}}
\begin{proof}
We can compute $\graph{2}$ from $\graph{1}$ in $O(m)$ time. Additionally, if in time $O(T(m))$, we have $\rpoly_{\graph{\ell}}^3(\prob,\dots,\prob)$ for $\ell\in [2]$, then the theorem follows by \Cref{lem:lin-sys}.
\qed
\end{proof}
In other words, if \Cref{th:single-p} holds, then so must \Cref{th:single-p-hard}.  

