\section*{Problem 2 - 281032}

Determine all pairs \((f, g)\) of functions \(f\) and \(g\), which are defined for all real numbers and satisfy the following conditions (1) to (4)!

(1) \(f\) is a quadratic function, in whose representation \(y=f(x)\) the coefficient of \(x^{2}\) is 1.

(2) For all real \(x\), \(f(x+1)=g(x)\).

(3) \(f\) has exactly one real root.

(4) It holds that \(g(5)=4\).
}

According to (1), the function \(f\) has the equation \(f(x)=x^{2}+b x+c\) with still unknown real numbers \(b\) and \(c\). According to (2),

\[
g(x)=f(x+1)=(x+1)^{2}+b(x+1)+c=x^{2}+(b+2) x+b+c+1
\]

For (3) to be satisfied, the term \(x^{2}+b x+c\) must be a perfect square, i.e.,

\[
0=x^{2}+b x+c=\left(x-\frac{b}{2}\right)^{2}-\frac{b^{2}}{4}+c \quad \Rightarrow \quad-\frac{b^{2}}{4}+c=0 \quad \Rightarrow \quad c=\frac{b^{2}}{4}
\]

Substituting (6) into (5) and then accordingly (4) for \(x=5\) and the function value \(g(x)=4\) yields

\[
\begin{aligned}
g(x) & =x^{2}+(b+2) x+\frac{b^{2}}{4}+b+1 \\
g(5) & =5^{2}+(b+2) 5+\frac{b^{2}}{4}+b+1 \\
4 & =\frac{1}{4}\left(b^{2}+24 b+144\right)
\end{aligned}
\]

The last equation has the solutions \(b_{1}=-16\) and \(b_{2}=-8\). Back-substitution yields \(c_{1}=64\) and \(c_{2}=16\) with the 2 solutions

\[
\text { 1) } \quad f(x)=x^{2}-16 x+64 \quad ; \quad g(x)=x^{2}-14 x+49
\]

\[
\text { 2) } \quad f(x)=x^{2}-8 x+16 \quad ; \quad g(x)=x^{2}-6 x+9
\]

The verification confirms the solutions.