3. Find the polynomial $f(x)$ that satisfies $x f(x-1)=(x-98) f(x)$.

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3. Substituting $x=0,1, \cdots, 97$ into $x f(x-1)=(x-98) f(x)$, we know that $x=0,1, \cdots, 97$ are all roots of $f(x)$. Let $f(x)=x(x-1) \cdots(x-97) g(x)$, then we have
$$
x(x-1) \cdots(x-98) g(x-1) \equiv(x-98) x(x-1) \cdots(x-97) g(x),
$$

i.e., $g(x) \equiv g(x-1)$.
Next, we prove that $g(x) \equiv a$ (a constant). Otherwise, suppose deg $g = n > 0$, then it has $n$ complex roots $x_{1}, x_{2}, \cdots, x_{n}$. Thus, $g\left(x_{i}-1\right)=g\left(x_{i}\right)=0,1 \leqslant i \leqslant n$. Therefore, $x_{1}-1, x_{2}-1, \cdots, x_{n}-1$ are also $n$ roots of $g(x)$. But $\sum_{i=1}^{n} x_{i} \neq \sum_{i=1}^{n}\left(x_{i}-1\right)$, which contradicts Vieta's formulas.
Therefore, $f(x)=a x(x-1) \cdots(x-97)$, where $a$ is a constant.