Find the polynomials $P$ such that for all real $x:(x-16) P(2 x)=16(x-$ 1) $P(x)$.

## Dessert



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With $x=1$, we have $P(2)=0$. Then $x=2$ gives $P(4)=0$. Similarly, $P(8)=P(16)=$ 0. We can thus write $P(X)=(X-2)(X-4)(X-8)(X-16) Q(X)$. By substituting back into the equation, we find $Q(x)=Q(2 x)$ for all $x$ (except $x=2,4,8,16$), so $Q(X)-Q(2 X)$ is the zero polynomial.

If $r \neq 0$ is a root of $Q$, then $2^{n} r$ is also a root for all $n \in \mathbb{N}$, so $Q=0$ or $r=0$. Therefore, $Q(X)=a X^{k}$, and it is easy to see that $k=0$. Thus, $Q$ is constant.

We deduce that $P=c(X-2)(X-4)(X-8)(X-16)$ with $c$ a real number. Conversely, this type of polynomial is indeed a solution.

## Dessert