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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?

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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.

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  • $\begingroup$ Sorry for a dumb question: why is your first inequality true? $\endgroup$ Commented Oct 24, 2014 at 14:48
  • $\begingroup$ Ramiro, you're right, this inequality is false. The LHS is at least c, which can be chosen arbitrarily large, in particular greater than (a+b)+1. $\endgroup$
    – Visitor
    Commented Oct 24, 2014 at 15:00
  • $\begingroup$ @Ramiro: for integers $c,d \geq 2$, $cd \geq c+d$. $\endgroup$ Commented Oct 24, 2014 at 16:08
  • $\begingroup$ @ZackWolske: I still don't see it. $\endgroup$ Commented Oct 24, 2014 at 16:37
  • $\begingroup$ @Visitor: You can't choose c greater than (a+b)+1, because you are constrained by the fact that $a+b = cd$, which in particular implies that $c \leq a+b$. $\endgroup$ Commented Oct 24, 2014 at 18:01