\section*{Problem 5 - 051035}

Provide conditions for the real numbers \(a, b, c, d\) that achieve the following:

1. If the conditions are satisfied, then the equation

\[
\frac{a(x+1)+b}{c(x+1)+d}=\frac{a x+b}{c x+d}
\]

(has at least) one solution.

2. If the equation (1) has a solution, then the conditions are satisfied.

Determine, if the conditions are satisfied, all solutions of (1). (Discussion)
}

If equation (1) has a solution \(x_{0}\), then it follows that:

\[
\begin{aligned}
& \frac{a\left(x_{0}+1\right)+b}{c\left(x_{0}+1\right)+d}=\frac{a x_{0}+b}{c x_{0}+d} \\
& c\left(x_{0}+1\right)+d \neq 0 \quad ; \quad c x_{0}+d \neq 0
\end{aligned}
\]

This implies

\[
\left(a\left(x_{0}+1\right)+b\right)\left(c x_{0}+d\right)=\left(c\left(x_{0}+1\right)+d\right)\left(a x_{0}+b\right)
\]

\(c^{2}+d^{2}>0\), because \(c\) and \(d\) cannot both be zero. And further

\[
a d=b c \quad ; \quad c^{2}+d^{2}>0
\]

If (1) has a solution, then (3) must hold, and conversely, if (3) holds, then (1) has a solution; because from (3) it follows that for all real numbers \(x_{0}\) (2) is certainly true, and from this it follows further:

I. If \(c=0\), then because of (3) \(a=0\) and \(d \neq 0\). Therefore, for every real number \(x_{0}\), (1) is certainly satisfied.

II. If \(c \neq 0\), then equation (1) is satisfied for every real number \(x_{0}\) with \(x_{0} \neq-\frac{d}{c}\) and \(x_{0} \neq-\frac{d+c}{c}\). \(x_{0}=-\frac{d}{c}\) and \(x_{0}=-\frac{d+c}{c}\) are not solutions of (1).

Adapted from \([5]\)