\section*{Problem 2 - 291042}

For two real numbers, it is required that:

The sum of the reciprocals of the two numbers and the reciprocal of the product of the two numbers is 1.

Determine all the values that can result as the sum \(s\) of two such numbers.
}

(I) For the sum \(s:=x+y\) of two real numbers \(x, y\) with

\[
\frac{1}{x}+\frac{1}{y}+\frac{1}{x y}=1
\]

it follows initially that

\[
\frac{1}{x}+\frac{1}{s-x}+\frac{1}{x(s-x)}=1
\]

and from this

\[
x^{2}-s x+s+1=0
\]

Since in (1) necessarily \(x \neq 0\) and \(y=s-x \neq 0\), it follows from (2) further that

\[
0 \neq x(s-x)=s x-x^{2}=s+1 \quad \text { i.e. } \quad s \neq-1
\]

On the other hand, (2) becomes through completing the square

\[
\left(x-\frac{s}{2}\right)^{2}=\frac{s^{2}}{4}-s-1
\]

Consequently, \(s^{2}-4 s-4 \geq 0,(s-2)^{2} \geq 8\) and finally

\[
|s-2| \geq 2 \sqrt{2}
\]

Overall, \(s\) thus necessarily lies in one of the intervals

\[
s<-1 ; \quad-1<s \leq 2-2 \sqrt{2} ; \quad s \geq 2+2 \sqrt{2}
\]

(II) Conversely, if \(s\) is from one of these intervals, then the numbers

\[
x=\frac{s}{2}+\sqrt{\frac{s^{2}}{4}-s-1} \quad \text { and } \quad y=\frac{s}{2}-\sqrt{\frac{s^{2}}{4}-s-1}
\]

are real (due to (4)), \(\neq 0\) (due to (3)), and it holds that \(x+y=s\) as well as \(\frac{1}{x}+\frac{1}{y}+\frac{1}{x y}=1\). Therefore, the sought values for \(s\) are precisely those that lie in one of the intervals from (5).