\section*{Problem 4 - 101234}

All polynomial functions \(y=f(x)\) are to be specified that satisfy the equations \(f(t \cdot x)=t \cdot f(x)\) for all real \(x\). Here, \(t\) is an arbitrarily given and then fixed real number.
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If \(t=1\), then apparently all polynomial functions \(f\) satisfy the property \(f(1 \cdot x)=f(x)=1 \cdot f(x)\). If \(t=0\), then it follows that \(f(0)=f(0 \cdot x)=0 \cdot f(x)=0\) for any real \(x\). This is clearly satisfied by all polynomial functions with a constant term of 0.

Now let \(t \notin \{0, 1\}\). Then the powers \(t, t^{2}, t^{3}, \ldots\) are all pairwise distinct. It follows inductively for all positive integers \(k\) that

\[
f\left(t^{k}\right)=f\left(t \cdot t^{k-1}\right)=t \cdot f\left(t^{k-1}\right)=\cdots=t^{k} \cdot f(1)
\]

We consider the polynomial function \(g(x):=f(x)-x \cdot f(1)\).

Then, with the property just shown, for all positive integers \(k\) the equation \(g\left(t^{k}\right)=f\left(t^{k}\right)-t^{k} \cdot f(1)=0\) holds, so \(g\) has the infinitely many distinct zeros \(t^{k}, k \in \mathbb{N}\). Since only one polynomial function, namely the zero function, has infinitely many zeros, it follows that \(g(x)=0\) for all \(x\), which means \(f(x)=x \cdot f(1)\) for all real numbers \(x\).

The only functions that satisfy this are the linear functions without a constant term, i.e., \(f(x)=a \cdot x\) with a real number \(a\). (Then \(f(1)=a\).) The verification confirms that these functions indeed satisfy the functional equation: \(f(t \cdot x)=a \cdot(t \cdot x)=t \cdot(a \cdot x)=t \cdot f(x)\).