300. Find all polynomials $P(x)$ for which the identity

$$
x P(x-1) \equiv (x-26) P(x)
$$

holds.
300. From the identity given in the problem, it follows that the sought polynomial $P(x)=P_{n}(x)$ (where $n$ is the degree of the polynomial) is divisible by $x$, i.e., $P_{n}(x)=x P_{n-1}(x)$, where $P_{n-1}(x)$ is some new polynomial of degree $n-1$. Therefore,

$$
P(x-1)=(x-1) P_{n-1}(x-1)
$$

and, hence,

$$
x(x-1) P_{n-1}(x-1)=x P(x-1)=(x-26) P(x),
$$

which implies that $P(x)$ is also divisible by $x-1$, i.e., $P_{n}(x)=x(x-1) P_{n-2}(x)$ (since $P_{n-1}(x)=(x-1) P_{n-2}(x)$ is divisible by $x-1$). In this case, $P(x-1)=(x-1)(x-2) P_{n-2}(x-1)$, and we obtain:

$x(x-1)(x-2) P_{n-2}(x-1)=$

$$
=(x-26) P_{n}(x)=(x-26) x(x-1) P_{n-2}(x),
$$

- from which it follows that $P_{n}(x)$ is also divisible by $x-2$ (i.e., $P_{n-2}(x)$ is divisible by $x-2$), and, therefore, $P_{n}(x)=x(x-1)(x-2) P_{n-3}(x)$. Substituting this value of $P(x)$ into the original relation, we similarly verify that $P(x)$ is also divisible by $x-3$, i.e., $P_{n}(x)=$ $=x(x-1)(x-2)(x-3) P_{n-3}(x)$, and so on.

Continuing in this manner, we eventually arrive at the following form of the polynomial $P(x)$:

$$
P(x)=P_{n}(x)=x(x-1)(x-2)(x-3) \ldots(x-25) P_{n-26}(x) .
$$

Substituting this expression for $P(x)$ into the given identity, we get:

$$
\begin{aligned}
& x(x-1)(x-2) \ldots(x-26) P_{n-26}(x-1)= \\
&=(x-26) x(x-1) \ldots(x-25) P_{n-26}(x),
\end{aligned}
$$

from which it follows that for the polynomial $P_{n-26}(x) \equiv Q(x)$ of degree $n-26$

$$
Q(x-1) \equiv Q(x)
$$

It is clear that if $Q(x)=Q_{0}(x)=c$ is a polynomial of degree zero (a constant!), then the relation (*) holds; we will show that it holds only in this case. Indeed, if $Q(x)=$ $=Q_{k}(x)=a_{0} x^{k}+a_{1} x^{k-1}+\ldots+a_{k-1} x+a_{k}, \quad$ where $k \geqslant 1$ and $a_{0} \neq 0$, then equating the coefficients of $x^{k-1}$ on both sides of the identity ( ${ }^{2}$ )

$$
a_{0}(x-1)^{k}+a_{1}(x-1)^{k-1}+\ldots=a_{0} x^{k}+a_{1} x^{k-1}+\ldots
$$

using the binomial theorem, we obtain:

$$
k a_{0}+a_{1}=a_{1}, \quad \text { or } \quad a_{0}=0,
$$

which, however, contradicts the assumption $a_{0} \neq 0$. Thus, $k=0, Q(x) \equiv c$ and

$$
P(x)=c x(x-1)(x-2) \ldots(x-25)
$$

- a polynomial of degree 26.