Prove or Disprove There Exist Only One Pair of Four Positive Integers whose sum equals their product.

Question

I have the following proof:

"We know that $$1+2+3=1\cdot 2\cdot 3.$$ Prove or Disprove: There exist four positive integers whose sum equals their product."

Now this seems pretty obvious at first that it can never exists. But the proposition also never states the the integers on the let hand side have to match the integers on the right hand side. With that said, how can that be used to help me solve this proof. I was thinking about using Minimum Counterexample but am unclear how to start.

Answer 1

To complete the task, we shall prove that the only solution is $1,1,2,4$.

First, we shall prove the following:

Assume that $x,y$ are real numbers such that $y\ge x\ge 2$. Then $xy\ge x+y$ and if $xy=x+y$ then $x=y=2$.

Proof:
$xy\ge 2y=y+y\ge x+y$.
If $x>2$ then $xy>2y\ge x+y$.
If $y>x$ then $xy\ge 2y>x+y$.

Now assume that $abcd=a+b+c+d$ and rename the numbers so that $a\le b\le c\le d$.

Suppose that $a\ge 2$. Since $2,2,2,2$ is not a solution, $d>2$.

Now we have that $abcd>abc+d$. Since $bc>a\ge 2$, $abc>a+bc\ge a+b+c$. Then $abcd>a+b+c+d$.

We get a contradiction. Then, $a=1$.

Now we have $bcd=1+b+c+d$. Suppose now that $b\ge 2$. Since $1,2,2,2$ is not a solution, $d> 2$. Similarly, we see that $bcd>bc+d\ge b+c+d$. So $b=1$.

The equation is now: $$cd=2+c+d$$ so $$c=\frac{d+2}{d-1}=1+\frac3{d-1}$$ Since $c$ is integer, $d-1$ must be $1$ or $3$. For $d-1=1$ we get the solution $1,1,4,2$ and for $d-1=3$ we get $1,1,2,4$.