\section*{Problem 2 - 171232}

For every integer \(a\), determine all real solutions \(x\) of the equation

\[
x^{4}+x^{3}+a^{2} x^{2}+x+1=0
\]
}

The equation can be transformed as follows:

\[
x^{4}+x^{3}+a^{2} x^{2}+x+1=x^{3} \cdot(x+1)+a^{2} x^{2}+(x+1)=\left(x^{3}+1\right) \cdot(x+1)+a^{2} x^{2}
\]

\((x+1)\) is negative if and only if \(x\) is less than -1 \((x+1<0 \Leftrightarrow x<-1)\).

\(\left(x^{3}+1\right)\) is also negative if and only if \(x\) is less than -1 \(\left(x^{3}+1<0 \Leftrightarrow x^{3}<-1 \Leftrightarrow x<-1\right)\).

Therefore, \((x+1)\) and \(\left(x^{3}+1\right)\) are always both negative or both greater than or equal to 0. Thus, their product \((x+1) \cdot\left(x^{3}+1\right)\) is always greater than or equal to 0.

\(a^{2} x^{2}\) is always greater than or equal to 0 as a square number.

Therefore, \(\left(x^{3}+1\right) \cdot(x+1)+a^{2} x^{2} \geq 0\), where the equality case only occurs if both \(\left(x^{3}+1\right) \cdot(x+1)\) and \(a^{2} x^{2}\) are equal to 0.

Thus, the following must hold:

(1) \(\left(x^{3}+1\right) \cdot(x+1)=0 \Leftrightarrow x=-1\) and

(2) \(a^{2} x^{2}=0 \Leftrightarrow a=0\) or \(x=0\). From \(x=-1\) (as mentioned above), it follows that \(a=0\).

Therefore, there is only a solution for \(a=0\), which is \(x=-1\).