# Can $p(x)q(x+1)$ and $q(x)p(x+1)$ be the same?

Suppose $$p(x)$$ and $$q(x)$$ are distinct polynomials with integer coefficients. Can we conclude that $$p(x)q(x+1)$$ and $$q(x)p(x+1)$$ are distinct expressions?

• Consider $f(x):=p(x)/q(x).$ What do we know about $f(x)$ if $p(x)q(x+1)=q(x)p(x+1)$? – Somos Apr 9 at 3:50
• Building on @Somos' hint. Look at $f(x)=p(x)/q(x)$. We can cancel comomon factors. You can prove, by induction that $f(x)=f(x+n)$ for all $n$. Then as in an old answer of mine it follows that either $p$ and $q$ have a common zero (and hence still a common factor), or infinitely many zeros. Both of which are absurd. Observe that the conclusion does not hold in prime characteristic. – Jyrki Lahtonen Apr 9 at 5:01
• @All Thank you Somos & Jyrki for your comments. I've included details of them in my answer, along with recognizing they came from you. – John Omielan Apr 9 at 5:23
• @JyrkiLahtonen You can also use calculus. With the rational function $f(x) = p(x)/q(x)$, if $q(x) \neq 0$, then it has a finite # of zeros. Thus, there are an infinite # of $x \in \mathbb{Z}$ such that $f(x)$ is continuous and differentiable between $x$ & $x+1$. Since $f(x) = f(x+1)$, by the Mean Value Theorem, there is a $c$ between $x$ & $x+1$ such that $f'(c) = 0$, i.e., it has an infinite # of zeros. However, $f'(x)$ is also a rational function, so it can only have a finite # of zeros unless it's $0$ itself. In that case, $f'(x) = 0$ means $f(x) = c$ for some constant $c$. – John Omielan Apr 9 at 18:06

## 3 Answers

No, you can't conclude that they're necessarily distinct expressions. An extremely simple example is if $$p(x) = a$$ and $$q(x) = b$$ where $$a$$ and $$b$$ are different integer constants, so then $$p(x)q(x + 1) = q(x)p(x + 1) = ab$$. A non-constant polynomial example is to let $$p(x) = x$$ and $$q(x) = -x$$. Then $$p(x)q(x + 1) = x(-(x + 1)) = -x(x + 1)$$ and $$q(x)p(x + 1) = -x(x + 1)$$.

As Somos suggested in a question comment, consider that $$q(x) \neq 0$$ so can use the rational function $$f(x) = \frac{p(x)}{q(x)}$$. Then, dividing both sides of $$p(x)q(x + 1) = q(x)p(x + 1)$$ by $$q(x)q(x + 1)$$ gives $$\frac{p(x)}{q(x)} = \frac{p(x+1)}{q(x+1)}$$, i.e., $$f(x) = f(x + 1)$$. For this to be true for all $$x$$ requires that $$f(x)$$ be a constant function (note that, as Jyrki Lahtonen wrote in a question comment, he gives a proof of this in his answer at Intersection of two subfields of the Rational Function Field in characteristic $$0$$), i.e., $$f(x) = c$$ so $$p(x) = cq(x)$$ for some rational constant $$c \neq 1$$ such that all of the coefficients in $$p(x)$$ are integers. In my $$2$$ examples above, $$c = \dfrac{a}{b}$$ (with $$b \neq 0$$) and $$c = -1$$. In the special case of $$q(x) = 0$$, then we have $$q(x) = cp(x)$$ instead, with $$c = 0$$.

• If $p$ is not $\pm q$, then can we conclude $p(x)q(x+1)$ and $q(x)p(x+1)$ are not the same? Does this have to do with the fact that $\mathbb{Z}[x]$ is a unique factorization domain? – user500144 Apr 9 at 3:16
• @user500144 I just realized a very example is for $p(x)$ and $q(x)$ to be integer constants as that is technically being polynomials (of $0$ degree). Of course, this provides an example of $p(x) \neq \pm q(x)$. I assume you want to have both $p(x)$ and $q(x)$ be of at least degree $1$. – John Omielan Apr 9 at 3:40
• +1 . I have posted a general answer for real polynomials. And I see you have added a lot to your answer while I was doing so. – DanielWainfleet Apr 9 at 5:42
• Thank you all for your answers. I was originally trying to prove $\mathbb{Q}$ is the fixed field of $\langle \phi\rangle$, where $\phi$ is the $\mathbb{Q}$-automorphism of $\mathbb{Q}(\pi)$ determined by $\pi\mapsto\pi+1$. So if we had a non-rational element $\alpha\in\mathbb{Q}(\pi)$ then we can express it as $\frac{p(\pi)}{q(\pi)}$, ($p,q$ are distinct polynomial expressions in $\pi$) and I was trying to determine that $\frac{p(\phi(\pi))}{q(\phi(\pi))}$ must be different from $\alpha$. So I guess in this case I can conclude it's different from $\alpha$, or is it a slightly different matter? – user500144 Apr 9 at 6:04
• @user500144 You are welcome for my help. However, I'm sorry but I'm not very familiar with the various terminology you are using, so I don't feel confident giving you an answer to your comment question. However, I believe one of the other people commenting or answering here can help you instead. – John Omielan Apr 9 at 6:13

We can prove the following theorem which essentially answers your question:

If $$p(x)$$ and $$q(x)$$ are distinct monic polynomials with complex coefficients and $$p(x)q(x+1)=q(x)p(x+1)$$ then $$p=q$$.

This has an easy converse too:

If $$p(x)$$ and $$q(x)$$ have either that $$p=cq$$ or $$q=cp$$ for some complex $$c$$, then $$p(x)q(x+1)=q(x)p(x+1)$$

Together, these classify entirely the solutions to your equation.

To prove the offered theorem, we first note that if a pair $$p(x)$$ and $$q(x)$$ have this property and $$r(x)$$ is any polynomial dividing both, then $$\frac{p(x)}{r(x)}$$ and $$\frac{q(x)}{r(x)}$$ also must have the given property, as both sides are divided by $$r(x)r(x+1)$$. Thus, if there is a counterexample, there is a counterexample where $$p(x)$$ and $$q(x)$$ are coprime. We assume now that $$p(x)$$ and $$q(x)$$ are coprime, so share no roots.

Let $$r$$ be a root of $$p(x)q(x+1)=q(x)p(x+1)$$ with maximal real part. Note that $$r$$ cannot be a root of $$q(x+1)$$, as then $$r+1$$ is a root of $$q(x)$$. Similarly, $$r$$ is not a root of $$p(x+1)$$. Thus, $$r$$ must be a root of both $$p(x)$$ and $$q(x)$$. This contradicts coprimality. Thus, to the contrary, $$p(x)q(x+1)=q(x)p(x+1)$$ must have no roots so $$p(x)=q(x)=1$$.

(1). Suppose $$p,q$$ are real polynomials with $$deg(p)\ge deg(q)>0$$ and $$\forall x\in \Bbb R\,(\,p(x)q(x+1)=p(x+1)q(x)\,).$$

We have $$p=qr+s$$ where $$r,s$$ are polynomials with $$deg(s)

Let $$A=\{x\in \Bbb R: \forall x\in \Bbb N \,q(x+n)\ne 0\}.$$ Observe that $$A$$ is an infinite set (since the polynomial $$q$$ is not the constant $$0\,$$) a fact to be used later.

For $$x\in A$$ let $$B(x)=\frac {p(x)}{q(x)}.$$ Then for $$x\in A$$ we have $$B(x)=B(x+n)$$ for all $$n\in \Bbb N,$$ so $$B(x)=\lim_{n\to \infty;n\in \Bbb N} B(x+n)=$$ $$=\lim_{n\to \infty;n\in \Bbb N} r(x+n) +\frac {s(x+n)}{q(x+n)}.$$ Now $$\lim_{n\to \infty}\frac {s(x+n)}{q(x+n)}=0$$ because $$deg (q)>deg (s).$$ Also $$\lim_{n\to \infty}r(x+n)$$ cannot exist for the polynomial $$r$$ unless $$r$$ is a constant. So let $$r(y)=K\in \Bbb R$$ for all $$y\in \Bbb R.$$ Now we have $$\forall x\in A\;( B(x)=K).$$ Therefore $$\forall x\in A\,(p(x)-Kq(x)=0).$$ So the polynomial $$p-Kq$$ is $$0$$ on the infinite set $$A.$$ Therefore $$\forall x\in \Bbb R\;(p(x)=Kq(x)).$$

Now observe that the converse $$(p=Kq \land deg(p)\ge deg(q)>0)$$ implies $$\forall x\in \Bbb R \,(p(x)q(x+1)=p(x+1)q(x)\,).$$

Regarding constant polynomials: If $$q=0$$ then (obviously) $$p(x)q(x+1)=p(x+1)q(x).$$ If $$q$$ is a non-$$0$$ constant then $$\forall x \in \Bbb R(\,p(x)q(x+1)=p(x+1)q(x)\,)\implies \forall x\in \Bbb R\,(p(x+1)=p(x)\,)$$ which implies the polynomial $$p$$ is constant.