Determine the polynomials $p(x)$ satisfying $x\cdot p(x-1) = (x-26)\cdot p(x)$

Question

Determine the polynomials $p(x)$ satisfying $x\cdot p(x-1) = (x-26)\cdot p(x)$.

My Solution: Put $x=0$, we get $p(0) = 0$, Similarly put $x=26,$ we get $p(26) = 0$.

That means $x=0,26$ are two roots of $p(x)$.

So we can write $p(x) = c\cdot x\cdot (x-26)\cdot q(x)$, where $q(x)$ is a some other polynomial in $x$.

Now please help me, how can I proceed further, thanks in advance.

Answer 1

It is easy to see that the polynomial $p$ has roots $0,1,...,25$ and therefore

$p(x) =x(x-1)...(x-25)g(x)$,

where $g(x)$ is also polynomial.

Replacing the original equality we obtain that $g(x-1)= g(x)$. Now noting $g(x)-g(0)=h(x)$ polinomial with degree $n$, we find that $h$ is roots $0,1,...,n$ ie having more root than degree $n$. Hence $g (x) = c$, $c$ constant. Finally

$p(x)=cx(x-1)...(x-25).$

Answer 2

Additional hint (after you correct your off-by-one-error): If $x$ is a (possibly complex) nonzero root of $p$, then $x-1$ is also a root.