\section*{Problem 4 - 101044}

Determine all quadratic functions \(f(x)\) that satisfy the equation \(f(x+1)=f(-x)\) for all real \(x\).
Let \(f(x)\) be the quadratic function \(f(x)=a x^{2}+b x+c\) with \(a \neq 0\). Then the condition in the problem statement is equivalent to the statement that for all real numbers \(x\) the equation

\[
\begin{gathered}
a(x+1)^{2}+b(x+1)+c=a \cdot(-x)^{2}+b \cdot(-x)+c \quad \text { i.e., } \\
a x^{2}+(2 a+b) x+(a+b+c)=a x^{2}+(-b) \cdot x+c
\end{gathered}
\]
or \((2 a+2 b) x+(a+b)=0\). Since this must hold for all real numbers \(x\), it must be that \(2 a+2 b=0\), so \(b=-a\).

Indeed, all quadratic functions \(f(x)=a x^{2}-a x+c\) with arbitrary real numbers \(a \neq 0\) and \(c\) satisfy the desired equation, because

\[
\begin{aligned}
f(x+1)=a(x+1)^{2}-a(x+1) & +c=a(x+1)(x+1-1)+c=a(x+1) x+c=a\left(x^{2}+x\right)+c= \\
& =a(-x)^{2}-a(-x)+c=f(-x)
\end{aligned}
\]