Find all polynomials $P(x)$ for which the identity holds: $x P(x-1) \equiv (x-26) P(x)$.

#
Substituting $x=0$, we get $0=-26 P(0)$, that is, $P(0)=0$. Substituting $x=1$, we get $P(0)=-25 P(1)$, that is, $P(1)=0$. Then we substitute $x=2,3, \ldots, 25$ and sequentially obtain $P(2)=P(3)=\ldots=P(25)=0$. Therefore, $P(x)=x(x-1)(x-2) \ldots(x-25) Q(x)$, where $Q(x)$ is some polynomial. From the identity $x P(x-1)=(x-$ 26) $P(x)$ it follows that $Q(x-1)=Q(x)$, that is, $Q(x)=c$ - a constant (see problem 61433).

## Answer

$P(x)=c x(x-1)(x-2) \ldots(x-25)$, where $c$ is some constant.