find all polynomials that satisfy $(x-16)P(2x)=16(x-1)P(x)$

Question

find all polynomials that satisfy this functional equation $$\forall x \in \mathbb{R} \\ (x-16)P(2x)=16(x-1)P(x)$$

I write the polynomial $$P(x)=\sum_{k=0}^n a_k x^k$$ but I found by comparing the LHS ans RHS (after rearrangement) that all coefficients must be zero, except the fourth coefficient; $P(x)=a x^4$. But when I re-substitute it into the equation I found that it's inconsistent but it make sense only if $a=0$.

Answer 1

If you use the brute force and write $$P(x)=\sum_{k=0}^n a_k\, x^k$$ $$ (x-16)P(2x)-16(x-1)P(x)=0$$ and solve for one sigle $a_k$ at the time, you will have succesively $$a_1=-\frac{15 a_0}{16}\qquad a_2=\frac{35 a_0}{128}\qquad a_3=-\frac{15 a_0}{512}\qquad a_4=\frac{a_0}{1024}$$ and, for $k>4$, $a_k=0$.

So $$P(x)=a_0\left(1-\frac{15 x}{16}+\frac{35 x^2}{128}-\frac{15 x^3}{512}+\frac{x^4}{1024} \right)$$ that is to say $$P(x)=\frac{a_0}{1024}(x-16) (x-8) (x-4) (x-2)$$

Answer 2

It is easy to see that the only constant solution is $0$. Now suppose that $P(x)$ is not a constant. We remark that, if $P(x)$ is a solution, so is $cP(x)$ for any constant $c$ so we may suppose that the leading term of $P(x)$ is $x^n$ for some $n$. Comparing the leading terms on the two sides we see that $n=4$.

Letting $x=1$ or $x=16$ shows us that $$P(2)=0=P(16)$$ Letting $x=2$ tells us that $P(4)=0$ and letting $x=4$ then tells us that $P(8)=0$. It then follows that the only candidate for a monic quartic solution would be $$P(x)=(x-2)(x-4)(x-8)(x-16)$$

It is now a simple matter to confirm that this polynomial does satisfy the functional equation, so the general solution is $$P(x)=c(x-2)(x-4)(x-8)(x-16)$$ for some constant $c$.

Answer 3

By plugging in $x=16$, we find $P(16)=0$. So $P$ has some positive roots. If $P$ is not $\equiv 0$, it has a minimal positive root $x_0$. Plug in $\frac 12x_0$ to find $0=16(\frac12x_0-1)P(\frac12x_0)$ and so $x_0=2$ as $P(\frac12x_0)\ne 0$. By plugging in $x=2$, we then find $P(4)=0$, next $P(8)=0$ and $P(16)=0$. Similar to above, we find that the maximal positive root is $16$. So $$P(x)=(x-2)(x-4)(x-8)(x-16)Q(x) $$ and therefore $$ (x-16)P(2x)=(x-16)(2x-2)(2x-4)(2x-8)(2x-16)Q(2x)=16(x-1)(x-2)(x-4)(x-8)(x-16)Q(2x)$$ whereas $$ 16(x-1)P(x)=16(x-1)(x-2)(x-4)(x-8)(x-16)Q(x).$$ Therefore $Q(2x)=Q(x)$ for all $x$ (except possibly for $x=1,2,4,8,16$). Hence $Q$ is bounded, hence constant.