\section*{Problem 2 - 091212}

a) All real solutions of the equation \(x(x+1)(x+2)(x+3)=\frac{9}{16}\) are to be determined.

b) Furthermore, all real numbers \(a\) are to be specified for which the equation \(x(x+1)(x+2)(x+3)=a\) has no, exactly one, exactly two, exactly three, exactly four, or more than four different real solutions in \(x\).
}

The general case b) is solved immediately:

b) If we set \(z=x+\frac{3}{2}\), then \(x\) is a solution of the equation

\[
x(x+1)(x+2)(x+3)=a
\]

if and only if \(z\) is a solution of the equation

\[
\begin{aligned}
\left(z-\frac{3}{2}\right)\left(z+\frac{3}{2}\right)\left(z-\frac{1}{2}\right)\left(z+\frac{1}{2}\right) & =a \quad \text { or } \\
\left(z^{2}-\frac{9}{4}\right)\left(z^{2}-\frac{1}{4}\right) & =a \\
z^{4}-\frac{5}{2} z^{2}+\frac{9}{16}-a & =0
\end{aligned}
\]

Now, equation (3) is satisfied if and only if

\[
z^{2}=\frac{5}{4}+\sqrt{\frac{25}{16}-\frac{9}{16}+a}=\frac{5}{4}+\sqrt{1+a} \quad \text { or } \quad z^{2}=\frac{5}{4}-\sqrt{1+a}
\]

We now distinguish the following cases:

1. Case: (also the answer to problem a):

For \(a=\frac{9}{16}\), it follows from (4) and (5) that

\[
z^{2}=\frac{5}{4}+\sqrt{1+\frac{9}{16}}=\frac{10}{4} \quad ; \quad z^{2}=\frac{5}{4}-\sqrt{1+\frac{9}{16}}=0
\]

Therefore, equation (3) has exactly three solutions in this case, namely

\[
z_{1}=0 ; \quad z_{2}=\frac{1}{2} \sqrt{10} ; \quad z_{3}=-\frac{1}{2} \sqrt{10}
\]

Thus, equation (1) also has exactly three solutions in this case, namely

\[
x_{1}=-\frac{3}{2} ; \quad x_{2}=-\frac{3}{2}+\frac{1}{2} \sqrt{10} ; \quad x_{3}=-\frac{3}{2}-\frac{1}{2} \sqrt{10}
\]

2. Case: For \(a-1\) and \(\sqrt{1+a}>\frac{9}{16}\), \(\frac{5}{4}-\sqrt{1+a}<\frac{9}{16}\), exactly three solutions if \(a=\frac{9}{16}\), exactly four solutions if \(-1<a<\frac{9}{16}\), more than four solutions in no case.

Adapted from \([5]\)