G4.3 Given two positive integers $x$ and $y, x y-(x+y)=\operatorname{HCF}(x, y)+\operatorname{LCM}(x, y)$, where $\operatorname{HCF}(x, y)$ and $\operatorname{LCM}(x, y)$ are respectively the greatest common divisor and the least common multiple of $x$ and $y$. If $c$ is the maximum possible value of $x+y$, find $c$.
Without loss of generality assume $x \geq y$.
Let the H.C.F. of $x$ and $y$ be $m$ and $x=m a, y=m b$ where the H.C.F. of $a, b$ is 1 .
L.C.M. of $x$ and $y=m a b . a \geq b$.
$$
\begin{array}{l}
x y-(x+y)=\mathrm{HCF}+\mathrm{LCM} \Rightarrow m^{2} a b-m(a+b)=m+m a b \\
a b(m-1)=a+b+1 \\
m-1=\frac{1}{a}+\frac{1}{b}+\frac{1}{a b} \\
1 \leq m-1=\frac{1}{a}+\frac{1}{b}+\frac{1}{a b} \leq 3 \\
m=2,3 \text { or } 4
\end{array}
$$
when $m=2, \frac{1}{a}+\frac{1}{b}+\frac{1}{a b}=1 \Rightarrow a+b+1=a b \Rightarrow a b-a-b-1=0$
$$
\begin{array}{l}
a b-a-b+1=2 \\
(a-1)(b-1)=2 \\
a=3, b=2, m=2, x=6, y=4, c=x+y=10
\end{array}
$$

When $m=3, \frac{1}{a}+\frac{1}{b}+\frac{1}{a b}=2 \Rightarrow a+b+1=2 a b \Rightarrow 2 a b-a-b-1=0$
$$
\begin{array}{l}
4 a b-2 a-2 b+1=3 \\
(2 a-1)(2 b-1)=3 \\
a=2, b=1, m=3, x=6, y=3, c=x+y=9
\end{array}
$$

When $m=4, \frac{1}{a}+\frac{1}{b}+\frac{1}{a b}=3 \Rightarrow a+b+1=3 a b \Rightarrow 3 a b-a-b-1=0$
$$
\begin{array}{l}
9 a b-3 a-3 b+1=4 \\
(3 a-1)(3 b-1)=4 \\
a=1, b=1, m=4, x=4, y=4, c=x+y=8 \\
\text { Maximum } c=10
\end{array}
$$

Maximum $c=10$