TITLE: $2 \times 3 = 5+1$ and $2+3 = 5 \times 1$. When else can we switch the operators like this?
QUESTION [5 upvotes]: I noticed the following:
$$2 \times 3 = 5+1$$
If you switch the operators, it is still true:
$$2+3 = 5 \times 1$$
There is another obvious/trivial example where you can swap the operators:
$$2\times 2 = 2+2$$
I think these are the only solutions (for positive integers). Can anyone give an elegant proof?

REPLY [5 votes]: Your claim is correct. Indeed, suppose that $ab = c+d$ and $a+b = cd$.
It is clear that at most one of the values $a,b,c,d$ can be equal to $1$, so w.l.o.g. we may assume that $a$ and $b$ are both at least $2$.
We have $$ab = c + \frac{a+b}{c} \leq (a+b)+1,$$
and dividing by $b$ gives
$$a \leq \frac{a}{b} + 1 + \frac{1}{b} < \frac{a}{2} +2,$$
from which we get $a < 4$; hence $a=2$ or $a=3$.
Similarly, $b=2$ or $b=3$.
The four remaining possibilities for $(a,b)$ can now be checked one by one.