What does this connection between addition and multiplication mean? I noticing that some numbers when multiplied together give the same product as their sum. After some trial and error I came up with a simple rule for this.
$$n\times\frac{n}{n-1} = n+\frac{n}{n-1}$$
For example
$$7\times\frac{7}{6} = \frac{49}{6}$$  $$ 7+\frac{7}{6} = \frac{49}{6}$$
I was wondering a few things,


*

*Is there any significance to this "theorem"?

*If it has already been discovered is there a name for it?

*Is there a similar theorem for division and subtraction?

 A: This is a consequence of the following: 
$$
n + \frac{n}{n - 1} = \frac{n(n -1)}{n - 1} + \frac{n}{n - 1} = \frac{n^2 - n + n}{n - 1} = \frac{n^2}{n - 1} = n\times \frac{n }{n - 1}.
$$
A: We have
$$\begin{align}
n + \frac{n}{n - 1} &= \frac{n(n -1)}{n - 1} + \frac{n}{n - 1} \\
&= \frac{n^2 - n + n}{n - 1} \\
&= \frac{n^2}{n - 1} \\
&= n\times \frac{n }{n - 1}.
\end{align}
$$
A: Notice $1 + \frac 1{n-1} = \frac {n-1}{n-1} + \frac 1{n-1} = \frac n{n-1}$
So $n\times \frac n{n-1} = n(1 + \frac 1{n-1}) = n + \frac n{n-1}$
Also note:  If we were asked to solve $m*n = m + n$ then $m*n -m = n$ and $m(n-1) = n$.  
Then if $n -1 = 0$ then $n = 1$ and $m*0 = 1$ which is impossible.  So $n\ne 1$ and $n -1 \ne 0$ and so $m = \frac n{n-1}$. 
So the only solutions are if $n \frac n{n-1}=n + \frac n{n-1}$ and we must solve for $n$.
If $n = 0$ then $n \frac n{n-1}=0$ and $n + \frac n{n-1} = 0$ so $n = 0$ is a solution.
If $n \ne 0$ then $n = \frac {n-1}n (n + \frac n{n-1}) = (n-1) + 1 = n$ so an number (integer or  not) will be a solution.  
So $nm = n + m \iff (n,m) = (0,0) $ or $m = \frac n{n-1}$ (and $m =  \frac n{n-1} \iff n = \frac m{m-1}$).
