7
$\begingroup$

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?

$\endgroup$
  • 4
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
5
$\begingroup$

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 $c = 0$.

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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 $p(x)=q(x)=1$.

$\endgroup$
1
$\begingroup$

(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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.