G4.4 If $x_{1}, x_{2}, x_{3}, x_{4}$ and $x_{5}$ are positive integers that satisfy $x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=x_{1} x_{2} x_{3} x_{4} x_{5}$, that is the sum is the product, find the maximum value of $x_{5}$.
The expression is symmetric. We may assume that $1 \leq x_{1} \leq x_{2} \leq x_{3} \leq x_{4} \leq x_{5}$.
If $x_{1}=x_{2}=x_{3}=x_{4}=1$, then $1+1+1+1+x_{5}=1 \times 1 \times 1 \times 1 \times x_{5} \Rightarrow$ no solution
$$
\begin{array}{l}
\therefore x_{1} x_{2} x_{3} x_{4}-1 \neq 0 \\
x_{1}+x_{2}+x_{3}+x_{4}=\left(x_{1} x_{2} x_{3} x_{4}-1\right) x_{5} \\
x_{5}=\frac{x_{1}+x_{2}+x_{3}+x_{4}}{x_{1} x_{2} x_{3} x_{4}-1}
\end{array}
$$

When $x_{5}$ attains the maximum value, the denominator must be 1 , i.e. $x_{1} x_{2} x_{3} x_{4}=2$
$$
\because 1 \leq x_{1} \leq x_{2} \leq x_{3} \leq x_{4} \therefore x_{1}=1, x_{2}=1, x_{3}=1, x_{4}=2 \text {, max. } x_{5}=\frac{1+1+1+2}{2-1}=5
$$

Method 2 We begin from the lowest integer.
Case 1 Let $x_{1}=x_{2}=x_{3}=x_{4}=1$, then $1+1+1+1+x_{5}=1 \times 1 \times 1 \times 1 \times x_{5} \Rightarrow$ no solution
Case 2 Let $x_{1}=x_{2}=x_{3}=1$ and $x_{4}>1$, then $3+x_{4}+x_{5}=x_{4} x_{5} \Rightarrow x_{5}=\frac{x_{4}+3}{x_{4}-1}$
When $x_{4}=2, x_{5}=5$; when $x_{4}=3, x_{5}=3$
When $x_{4}=4$, no integral solution for $x_{5}$
When $x_{4}=5, x_{5}=2$, contradicting that $1 \leq x_{1} \leq x_{2} \leq x_{3} \leq x_{4} \leq x_{5}$.
When $x_{4}>5$, then $x_{5}1$, then $2+x_{3}+x_{4}+x_{5}=x_{3} x_{4} x_{5}$
When $x_{3}=2,4+x_{4}+x_{5}=2 x_{4} x_{5} \Rightarrow x_{5}=\frac{x_{4}+4}{2 x_{4}-1}>1 \Rightarrow x_{4}+4>2 x_{4}-1 \Rightarrow x_{4}<5$
When $x_{4}=2, x_{5}=2$
When $x_{4}=3,4$, no integral solution for $x_{5}$

Case $41=x_{1}<x_{2} \leq x_{3} \leq x_{4} \leq x_{5}$, then $x_{5}=\frac{1+x_{2}+x_{3}+x_{4}}{x_{2} x_{3} x_{4}-1}<\frac{1+3 x_{4}}{4 x_{4}-1}$
When $x_{2}=x_{3}=x_{4}=2, x_{5}=1<x_{4}$, contradiction
When $2 \leq x_{2}=x_{3}<x_{4}$
$1+3 x_{4}<4 x_{4}-1$
$\frac{1+3 x_{4}}{4 x_{4}-1}<1 \Rightarrow x_{5}<1$, contradiction $\therefore$ There is no integral solution for $x_{5}$.

Case $52 \leq x_{1} \leq x_{2} \leq x_{3} \leq x_{4} \leq x_{5}$, then $x_{5}=\frac{x_{1}+x_{2}+x_{3}+x_{4}}{x_{1} x_{2} x_{3} x_{4}-1}<\frac{4 x_{4}}{8 x_{4}-1}$
$$
\begin{array}{l}
1<4 x_{4} \\
4 x_{4}<8 x_{4}-1
\end{array}
$$
$\frac{4 x_{4}}{8 x_{4}-1}<1 \Rightarrow x_{5}<1$, contradiction $\therefore$ There is no integral solution for $x_{5}$.
Conclusion: The solution set for $\left(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right)$ is $\{(1,1,1,2,5),(1,1,1,3,3),(1,1,2,2,2)\}$. Maximum for $x_{5}=5$