I was just relating some random equations to find out if I get something new, wherein, I tried this :
Let \begin{equation} E=p+q \end{equation} $ \text{ for some }p,q\in\mathbb{Z} ; p>q$
$p,q$ are non-consecutive.
Also, let$$\mathbb{E}=p!+q!$$
Then, I tried to relate the above equations. I wished to relate $E$ and $\mathbb{E}$.
I found this :
$\because p>q\implies$ at some point in $p!$, we will have \begin{align} p!=&p(p-1)(p-2)\cdots(p-(p-q-1))(p-(p-q))(p-(p-q+1))(p-(p-q+2))\cdots 2\times1. \\ \equiv&p(p-1)(p-2)\cdots(p-(p-q-1)) \ q! \end{align}
Now, from above,
\begin{align} p!+q!&=q!\Big(p(p-1)(p-2)\cdots(p-(p-q-1)) + 1\Big)\\ &\equiv q!\Big(p(p-1)(p-2)\cdots(q+1) + 1\Big)\\ \implies \mathbb{E}&=q!\Big(p(p-1)(p-2)\cdots(q+1) + 1\Big) \end{align}
Solving for $p\implies$
\begin{align} p&=\frac{\mathbb{E}-q!}{q! \ \Big((p-1)(p-2)\cdots(q+1)\Big)}\\ &\equiv \frac{\mathbb{E}-q!}{Q}\\ \end{align} $\text{where }Q=q! \ \Big((p-1)(p-2)\cdots(q+1)\Big)$
Now, solving for $q\implies$
\begin{align} q&=E-p\\ &=E-\frac{\mathbb{E}-q!}{Q}\\ &=\frac{EQ-(\mathbb{E}-q!)}{Q} \end{align}
Thus,
\begin{align} &p=\frac{\mathbb{E}-q!}{Q}\\ &q=\frac{EQ-\mathbb{E}+q!}{Q} \end{align} where $Q=q! \ \Big((p-1)(p-2)\cdots(q+1)\Big)$
Quick Check
Take \begin{align} p=5&, q=3\\ \implies &E=p+q=8\\ Q&=3!(4)=24\\ \end{align} From formula:$$\mathbb{E}=(8\times24)-(3\times24)+(3!)=126$$ From calculator:$$\mathbb{E}=5!=3!=126$$
Now, I can relate $E$ and $\mathbb{E}$ :)
However, I have a few problems :
1.The above approach took quiet a couple of hours for me to be figured out.
2.Also, in most cases, we need to use calculators.
So, I wish to ask if there is any alternate approach which minimises the labor to arrive at the consequence. Also, won't it be great if we get an approach, which requires minimal use of calculators !?
Any help in this regard is highly appreciated.
Thank You.
$Q$ is just $(p-1)!$, so, if I understand correctly, the "relation" you gave between $p!+q!$ and $p+q$ is just $$p!+q! = (p+q)(p-1)!-q(p-1)!+q!$$ which I'm afraid is not very interesting.