2k(k-1)(k-2)\dotsb(3)=k!, \end{align*} which is a contradiction. \end{proof} \begin{theorem}\label{semiprimes} There are no positive integer solutions to the Diophantine equation $$c_{n,k}=(pq)^\ell$$ with $3\leq k\leq n$, $\ell\geq2$, and distinct primes $p$ and $q$. \end{theorem} \begin{proof} First, consider the case $k=3$. Suppose that $$c_{n,3}=\frac{(n+3)(n+2)(n-2)}{6}=(pq)^\ell$$ for some $n,p,\ell\in\N$ such that $n\geq3$, $\ell\geq2$, and $p$ and $q$ are distinct primes. By computer exhaustion, we check that there are no integer solutions for $n\leq 122$. Consider $n>122$. Since $n+3$ and $n+2$ are greater than $6$ and are relatively prime, assume without loss of generality that $p\mid(n+3)$ and $q\mid(n+2)$. As $\gcd(n+3,n-2)\mid5$ and $\gcd(n+2,n-2)\mid4$, we have \begin{align*} (n+3,n+2,n-2)=&\hspace{2pt}(\alpha p^\ell,\beta q^\ell,\gamma),\ (\alpha 5^{\ell_1},\beta q^\ell,\gamma5^{\ell-\ell_1}),\\ &\hspace{2pt}(\alpha p^\ell,\beta2^{\ell_2},\gamma2^{\ell-\ell_2}),\text{ or }(\alpha5^{\ell_1},\beta2^{\ell_2},\gamma5^{\ell-\ell_1}2^{\ell-\ell_2}), \end{align*} where $\alpha,\beta,\gamma\in\N$, $\alpha\beta\gamma=6$, $\ell_1=1$ or $\ell-1$, and $\ell_2=1$, $2$, $\ell-2$, or $\ell-1$. Note that \begin{align*} \min\{\alpha p^\ell,\beta q^\ell,\gamma\}&\leq6,\\ \min\{\alpha 5^{\ell_1},\beta q^\ell,\gamma5^{\ell-\ell_1}\}&\leq6\cdot5=30,\\ \min\{\alpha p^\ell,\beta2^{\ell_2},\gamma2^{\ell-\ell_2}\}&\leq6\cdot4=24,\text{ and}\\ \min\{\alpha5^{\ell_1},\beta2^{\ell_2},\gamma5^{\ell-\ell_1}2^{\ell-\ell_2}\}&\leq6\cdot5\cdot4=120. \end{align*} All these violate the condition that $n>122$. Hence, no positive integer solutions exist when $k=3$. Next, consider the case $k>3$. Without loss of generality, let $p>q$. By Lemma~\ref{faulkner modified}, it follows that $p>k$. Let $T=\{n+k,n+k-1,\dotsc,n+2,n-k+1\}$. If $p^\ell\mid t_0$ for some $t_0\in T$, then $$(p^\ell-1)(p^\ell-2)\dotsb(p^\ell-k+2)(p^\ell-2k+1)\leq\prod_{t\in T\setminus\{t_0\}}t=\frac{p^\ell q^\ell k!}{t_0}\leq q^\ell\cdot k!.$$ Since $q

2$, depending on $\ell_1=1$ or $\ell-1$, we have either $$(p^{\ell-1}-1)(p^{\ell-1}-2)\dotsb(p^{\ell-1}-k+2)(p^{\ell-1}-2k+1)\leq\prod_{t\in T\setminus\{t_1\}}t=\frac{p^\ell q^\ell k!}{t_1}\leq pq^\ell\cdot k!$$ or $$(p^{\ell-1}+2k-1)(p^{\ell-1}+2k-2)\dotsb(p^{\ell-1}+k-1)\leq pq^\ell\cdot k!.$$ In either case, since $q^{\ell-1}\leq p^{\ell-1}-2k+1$, we always have $$(p^{\ell-1}-1)(p^{\ell-1}-2)\dotsb(p^{\ell-1}-k+2)\leq pq\cdot k!.$$ Furthermore, as $pq< p(p-1)

k$, we have
$$4!\cdot k(k-1)\dotsb5<4!\cdot(p^{\ell-1}-2)(p^{\ell-1}-3)\dotsb(p^{\ell-1}-k+3)