I was recently introduced to an interesting polynomial problem.
Problem. Determine all polynomials $p\left(x\right)\in \mathbb{R}\left[x\right]$ such that $\left(x – 16\right)p\left(2x\right) = 16\left(x – 1\right)p\left(x\right)$ for all $x\in \mathbb{R}$.
The way I solved this was to essentially use this equation as a feedback loop to peel factors off of $p$. Before we come to the solution, though, I would like to prove a quick lemma.
Lemma. If we have $a \in \mathbb{R}$, $a\neq\pm 1$, and $p\left(x\right)\in \mathbb{R}\left[x\right]$ such that $p\left(ax\right) = p\left(x\right)$ for all $x\in \mathbb{R}$, then $p$ is constant.
Proof.
For $a = 0$, we have $p\left(0\right)=p\left(x\right)$, and so $p$ is constant.
For $a\neq 0$, say $p\left(x\right) = c_{n}x^{n} + \cdots + c_{1}x + c_{0}$ for some $c_{i}\in \mathbb{R}$, $i = 0, \ldots, n$. Then, $p\left(ax\right) = p\left(x\right)$ for all $x\in \mathbb{R}$ gives $p\left(ax\right) – p\left(x\right) = 0$, and so $\left(a^{n} – 1\right)c_{n}x^{n} + \cdots + \left(a – 1\right)c_{1}x + c_{0} = 0$, which means that $c_{i} = 0$ for all $i>0$, since $a^{\ell} – 1\neq 0$ for all $\ell\geq 1$, leaving $p \equiv c_{0}$.
QED
We will be using the full lemma when we come to the theorem I prove later on, but the solution to the original problem will only need the case where $a = 2$.
Solution.
In the equation $\left(x – 16\right)p\left(2x\right) = 16\left(x – 1\right)p\left(x\right)$, let’s plug in $x = 1$. Then, we get
$$-15p\left(2\right) = 0 \implies p\left(2\right) = 0.$$
In other words, $x – 2\mid p\left(x\right),$ and so there exists a polynomial $q_{1}\left(x\right)$ such that $p\left(x\right) = \left(x – 2\right)q_{1}\left(x\right)$.
Plugging this new expression in, and factoring $2$ from the $\left(2x – 2\right)$ that would appear on the left-hand side, gives us $2\left(x – 16\right)\left(x – 1\right)q_{1}\left(2x\right) = 16\left(x – 1\right)\left(x – 2\right)q_{1}\left(x\right)$. Next, if we plug $2$ in to both sides, we find that there is a polynomial $q_{2}\left(x\right)$ such that $q_{1}\left(x\right) = \left(x – 4\right)q_{2}\left(x\right)$, and we repeat this process twice more to, eventually, get a polynomial $q_{4}\left(x\right)$ which, when plugged back in to the original equation, gives us
$$16\left(x – 16\right)\left(x – 1\right)\left(x – 2\right)\left(x – 4\right)\left(x – 8\right)q_{4}\left(2x\right) = 16\left(x – 1\right)\left(x – 2\right)\left(x – 4\right)\left(x – 8\right)\left(x – 16\right)q_{4}\left(x\right),$$
which holds for all $x\in \mathbb{R}$. Cancelling all of the common factors from both sides gives the equation $q_{4}\left(2x\right) = q_{4}\left(x\right)$, which, by the lemma above, means that $q_{4}\equiv C$ for some constant $C\in \mathbb{R}$.
Therefore, once we collect the factors backwards, back to our original goal of determining $p\left(x\right),$ we finally have that $p\left(x\right) = C\left(x – 2\right)\left(x – 4\right)\left(x – 8\right)\left(x – 16\right)$. $\blacksquare$
Note: The skeptical among you may have some reservations about my conclusion that $q_{4}\left(2x\right) = q_{4}\left(x\right),$ since the original equation does not necessarily imply that for $x = 1,2,4,8,16$, since those values of $x$ would result in dividing by $0$ when cancelling the factors out. However, we are okay: polynomials of degree $n$ are are uniquely determined by $n + 1$ points. So, the polynomial $q_{4}\left(2x\right) – q_{4}\left(x\right)$ being $0$ everywhere but $x = 1,2,4,8,16$ is enough to conclude that it would also be $0$ at those values, as well.
Now that we have that problem out of the way, it may be a good idea to look back at the original problem and find some patterns. After a bit of investigation, you’ll probably be able to quickly generalize the problem to the following:
Generalized Problem. Given $k\in\mathbb{N}$ and $a\in\mathbb{R}$, determine all polynomials $p\left(x\right)\in \mathbb{R}\left[x\right]$ such that $\left(x – a^{k}\right)p\left(ax\right) = a^{k}\left(x – 1\right)p\left(x\right)$ for all $x\in \mathbb{R}$.
We can verify that this does match the original problem’s pattern with $k = 2$ and $a = 2$. Now, I would like the reader to spend a few minutes testing a bunch of values of $k$ and $a$. Look for values of $k$ and $a$ that may break the pattern we saw in the original problem. Please do this before moving on, and try to formulate an answer to the generalized problem.
The following theorem resolves the generalized problem.
Theorem. Let $k\in \mathbb{N}$, $a\in \mathbb{R}$, and $p\left(x\right)\in \mathbb{R}\left[x\right]$. If $\left(x – a^{k}\right)p\left(ax\right) = a^{k}\left(x – 1\right)p\left(x\right)$ for all $x\in \mathbb{R}$, then
- If $a = 0$, then
- If $k \neq 0$, $p\left(x\right)$ is any polynomial with constant term $0$.
- If $k = 0$, then $p\left(x\right)$ is identically $0$.
- If $a = 1$, then there is no restriction on the polynomial.
- If $a = -1$, then
- If $k$ is even, then $p\left(x\right)$ is an even polynomial.
- If $k$ is odd, then $p\left(x\right)$ is of the form $\left(x + 1\right) q\left(x\right)$, where $q\left(x\right)$ is an odd polynomial.
- If $a\neq 0, \pm 1$, then there exists $C\in \mathbb{R}$ such that $p\left(x\right) = C\prod\limits_{1\leq j\leq k}\left(x – a^{j}\right)$.
In particular, if $k = 0$, then $p\left(x\right)$ is constant.
N.B. Whenever $k = 0$ in (4), the product will look like $\prod\limits_{1\leq j\leq 0}$. Since there is no number bigger than $1$ but less than $0$, this gives us what’s known as the “empty product,” which is always $1$, and so that’s why $p\left(x\right)$ is constant when $k = 0$ in that case.
Proof.
(1) When $a = 0$, then there are two subcases.
(A) If $k\neq 0$, the equation becomes $xp\left(0\right) = 0$, which is equivalent to saying $p\left(0\right) = 0$.
(B) If $k = 0$, then $0^{k} = 1$, and so the equation becomes $xp\left(0\right) = \left(x – 1\right) p\left(x\right) = xp\left(x\right) – p\left(x\right)$. Subtracting by $xp\left(0\right)$ gives $xp\left(x\right) – p\left(x\right) – xp\left(0\right) = 0$.
If we write $$p\left(x\right) = \sum\limits_{0\leq \ell\leq n} a_{\ell}x^{\ell},$$ then this equation becomes $$\sum\limits_{0\leq \ell\leq n} \left(a_{\ell}x^{\ell + 1} – a_{\ell}x^{\ell}\right) – xa_{0} = 0.$$ Hence, $a_{n} = 0$, since there is no other term of degree $n+1$ to cancel with, and $a_{\ell} = a_{\ell – 1}$ for all $\ell \geq 2$, since the degree $\ell\geq 2$ term in this expression is $\left(a_{\ell-1} – a_{\ell}\right)x^{j} = 0$. The degree $1$ term in this expression is $\left(a_{0} – a_{1} – a_{0}\right)x = -a_{1}x = 0$, so $a_{1} = 0$. But since $a_{\ell-1} = a_{\ell}$ for all $\ell\geq 2$, this means $a_{\ell} = 0$ for all $\ell\geq 1$. Therefore, of everything in the equation $$\sum\limits_{0\leq \ell\leq n} \left(a_{\ell}x^{\ell + 1} – a_{\ell}x^{\ell}\right) – xa_{0} = 0,$$ all that remains on the lefthand side is the constant term, corresponding to $\ell = 0$ in the sum, $-a_{0} = 0$, and so $a_{0} = 0$, making $p\left(x\right)$ identically $0$, as required.
(2) When $a = 1$, the equation becomes $\left(x – 1\right) p\left(x\right) = \left(x – 1\right) p\left(x\right),$ which reduces to $p\left(x\right) = p\left(x\right),$ which is satisfied by any polynomial.
(3) When $a = -1$, we have two subcases.
(A) If $k$ is even, the equation becomes $\left(x – 1\right) p\left(-x\right) = \left(x – 1\right) p\left(x\right),$ which reduces to $p\left(-x\right) = p\left(x\right),$ and so $p\left(x\right)$ must be an even polynomial.
(B) If $k$ is odd, the equation becomes $\left(x + 1\right) p\left(-x\right) = \left(x – 1\right) p\left(x\right).$ Plugging in $x = 1$ gives $2p\left(-1\right) = 0$, and so $p\left(x\right)$ is of the form $\left(x + 1\right) q\left(x\right)$. Plugging this back in to the original equation, and factoring out a $-1$ from the $\left(-x + 1\right)$ that will show up on the left-hand side, gives $-\left(x + 1\right) \left(x – 1\right) q\left(-x\right) = \left(x – 1\right) \left(x + 1\right) q\left(x\right).$ Cancelling the common factors gives $-q\left(-x\right) = q\left(x\right),$ and so $q\left(x\right)$ must be an odd function.
(4) When $a\neq 0, \pm 1$, we can finally use the full trick used in the solution of the original problem of this post. I’ll leave the full, formal induction proof to the reader and, instead, appeal to the obvious pattern that will arise when we do two steps of finding the factors of $p\left(x\right)$.
The equation we have is $$\left(x – a^{k}\right)p\left(ax\right) = a^{k}\left(x – 1\right)p\left(x\right).$$
Plugging in $x = 1$ gives $\left(1 – a^{k}\right) p\left(a\right) = 0,$ and so $p\left(x\right) = \left(x – a\right) q_{1}\left(x\right)$ for some polynomial $q_{1}\left(x\right)$. Plugging this back in, and factoring out the $a$ from the $\left(ax – a\right)$ that will show up on the left-hand side, gives the equation
$$a\left(x – a^{k}\right)\left(x – 1\right) q_{1}\left(ax\right) = a^{k}\left(x – 1\right)\left(x – a\right) q_{1}\left(x\right).$$
If we now plug in $x = a$, we get $a\left(a – a^{k}\right)\left(a – 1\right) q_{1}\left(a^{2}\right) = 0,$ which means that $q_{1}\left(x\right) = \left(x – a^{2}\right) q_{2}\left(x\right)$ for some polynomial $q_{2}\left(x\right).$
Here, I will leave the formal induction argument to the reader. Instead, I will just say: Iterate this process by plugging in $x = a^{j}$ at step $j$ to reveal that $\left(x – a^{j}\right)$ is a factor of $q_{j-1}\left(x\right)$, until we reach $j = k$ and the following equation:
$$a^{k} \left[\prod\limits_{0\leq j\leq k} \left(x – a^{j}\right)\right] q_{k}\left(ax\right) = a^{k} \left[\prod\limits_{0\leq j\leq k} \left(x – a^{j}\right)\right] q_{k}\left(x\right).$$
Here, we can finally cross out all of the common factors, leaving us simply with the equation $q_{k}\left(ax\right) = q_{k}\left(x\right),$ and so there exists $C\in \mathbb{R}$ such that $q_{k}\equiv C.$
Now, going backwards, recall that $q_{j-1}\left(x\right) = \left(x – a^{j}\right)q_{j}\left(x\right)$ for $j = 1,\ldots, k$ (where $q_{0}$ is just $p$). This recurrence gives us the following:
$$p\left(x\right) = C \prod\limits_{1\leq j\leq k} \left(x – a^{j}\right),$$
as required.
QED