\section*{Problem 4 - 281034}

Determine all pairs \((a ; b)\) of real numbers that satisfy the following equation (1)!

\[
a^{3}-b^{3}-a^{2}+b^{2}+a-b=0
\]
}

By polynomial division, it follows that \(a^{3}-b^{3}=(a-b)\left(a^{2}+a b+b^{2}\right)\) and by the binomial formula \(a^{2}-b^{2}=\)
\((a-b)(a+b)\). Then from \((1)\)

\[
\begin{aligned}
a^{3}-b^{3}-a^{2}+b^{2}+a-b & =(a-b)\left(a^{2}+a b+b^{2}\right)+(a-b)(a+b)+(a-b)=0 \\
& =(a-b)\left(a^{2}+a b+b^{2}+a+b+1\right)=0
\end{aligned}
\]

Obviously, all pairs \((a, b)\) with \(a=b\) satisfy equation (1). The second factor cannot be zero, since

\[
\begin{aligned}
a^{2}+a b+b^{2}+a+b+1 & =a^{2}+a(b+1)+b^{2}+b+1=0 \\
a_{1 ; 2} & =-\frac{b+1}{2} \pm \sqrt{\frac{b^{2}+2 b+1-\left(4 b^{2}+4 b+4\right)}{4}} \\
& =-\frac{b+1}{2} \pm \sqrt{-\frac{3}{4}\left(b^{2}+\frac{2}{3} b+1\right)}=-\frac{b+1}{2} \pm \sqrt{-\frac{3}{4}\left(\left(b+\frac{1}{3}\right)^{2}+\frac{8}{9}\right)}
\end{aligned}
\]

The radicand is thus always negative. Therefore, there is no real pair \((a, b)\) such that \(\left(a^{2}+a b+b^{2}+a+b+1\right)=0\). Equation (1) has only pairs \((a, b)\) with \(a=b\) as solutions.