\section*{Problem 3 - 191013}

Determine all real numbers \(x\) with \(-1 \leq x \leq 1\) for which the expression \(x^{2}+3 x+4\) is the square of a natural number!
}

For all real \(x\) it holds that

\[
x^{2}+3 x+4=\left(c+\frac{3}{2}\right)^{2}+\frac{7}{4}>1
\]

for \(-1 \leq x \leq 1\) it further holds that \(x^{2} \leq 1, 3 x \leq 3\), thus

\[
x^{2}+3 x+4 \leq 1+3+4<9
\]

From this it follows: If the term for real \(x\) with \(-1 \leq x \leq 1\) results in an integer square number, then this can only be the number 4.

The equation \(x^{2}+3 x+4=4\) is now equivalent to \(x(x+3)=0\).

Therefore, it has exactly the solutions \(x=0\) and \(x=-3\). Of these, only the first one satisfies the condition \(-1 \leq x \leq 1\). Thus, exactly the number \(x=0\) has the required property.