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?

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    $\begingroup$ Consider $f(x):=p(x)/q(x).$ What do we know about $f(x)$ if $p(x)q(x+1)=q(x)p(x+1)$? $\endgroup$ – Somos Apr 9 at 3:50
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    $\begingroup$ 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. $\endgroup$ – Jyrki Lahtonen Apr 9 at 5:01
  • $\begingroup$ @All Thank you Somos & Jyrki for your comments. I've included details of them in my answer, along with recognizing they came from you. $\endgroup$ – John Omielan Apr 9 at 5:23
  • $\begingroup$ @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 $c$. $\endgroup$ – John Omielan Apr 9 at 18:06

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 $c = 0$.

  • $\begingroup$ 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? $\endgroup$ – user500144 Apr 9 at 3:16
  • $\begingroup$ @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 $1$. $\endgroup$ – John Omielan Apr 9 at 3:40
  • $\begingroup$ +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. $\endgroup$ – DanielWainfleet Apr 9 at 5:42
  • $\begingroup$ 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? $\endgroup$ – user500144 Apr 9 at 6:04
  • $\begingroup$ @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. $\endgroup$ – 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 $deg(s)<deg(q).$

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.


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