$x_1+x_2+x_3+x_4+x_5=x_1x_2x_3x_4x_5$. $x_1,x_2,x_3,x_4,x_5$ are natural numbers. Find maximum of $x_1,x_2,x_3,x_4,x_5$

Question

$x_1+x_2+x_3+x_4+x_5=x_1x_2x_3x_4x_5$.

$x_1,x_2,x_3,x_4,x_5$ are natural numbers.

Find maximum of $x_1,x_2,x_3,x_4,x_5$.

Answer 1

Claim. Assume $x_1,\ldots,x_5\in\Bbb N$ with $x_1+x_2+x_3+x_4+x_5=x_1x_2x_3x_4x_5$. Then up to permutation we have $(x_1,x_2,x_3,x_4,x_5)=(1,1,1,2,5)$ or $(x_1,x_2,x_3,x_4,x_5)=(1,1,1,3,3)$. In particular, all $x_i$ are $\le 5$.

Proof. We may assume wlog. that $x_1\le\ldots\le x_5$. Then $x_1+\ldots+x_5\le 5x_5$ and we conclude $x_1x_2x_3x_4\le 5$. This leaves us only with the following possibilities for $(x_1,x_2,x_3,x_4)$: $$(1,1,1,1), (1,1,1,2), (1,1,1,3), (1,1,1,4), (1,1,1,5), (1,1,2,2).$$ These lead to the following conditions on $x_5$: $$\begin{align} 4+x_5=x_5&\implies\text{impossible}\\ 5+x_5=2x_5&\implies x_5=5\\ 6+x_5=3x_5&\implies x_5=3\\ 7+x_5=4x_5&\implies \text{impossible}\\ 8+x_5=5x_5&\implies \text{impossible}\\ 6+x_5=4x_5&\implies x_5=2\\ \end{align}$$ $\square$