# An equation of rational functions

I'm trying to get the set of solutions of the following equation, whose unknowns are the rational functions $$f$$ and $$g$$ :

$$\forall x\in\mathbb{R}$$ such that the LHS and RHS are both defined, $$f(x)f(x+1)=\frac{g(x+1)}{g(x)}$$

$$\left(f(x):=1, g(x):=1\right)$$ is an example of a trivial solution to the equation. Now I'm trying to characterize the entire set of solutions.

Since $$f$$ and $$g$$ are rational functions, we can write them as $$f(x)=\frac{P_1(x)}{Q_1(x)}$$ and $$g(x)=\frac{P_2(x)}{Q_2(x)}$$, with $$P_1, P_2, Q_1, Q_2\in\mathbb{R}[X]$$.

So this equation is equivalent to finding the set of polynomials $$P_1, P_2, Q_1, Q_2$$ such that :

$$\forall x,\hspace{6pt}\frac{P_1(x)P_1(x+1)}{Q_1(x)Q_1(x+1)}=\frac{P_2(x+1)Q_2(x)}{P_2(x)Q_2(x+1)}$$

I don't really have any ideas as of how to move forward...

Any suggestion ?

Edit: The answer has been modified following a comment of Harmonic Sun and to include a second, simpler proof.

Any solution is essentially of the form in Ivan Neretin's answer: $$f(x)=\pm\frac{R(x+1)}{R(x)},g(x)=\pm R(x+1)R(x),$$ where $$R(x)$$ is an arbitrary rational function. Observe that if a solution $$f,g$$ with complex coefficients is given we construct $$R$$ with complex coefficients such that $$f(x)=\pm\frac{R(x+1)}{R(x)},g(x)=R(x+1)R(x).$$

Proof: I will first work in the complex setting and then consider the real case. For a rational function $$h$$ and some complex $$\mu$$, we define $$v(h,\mu)=m$$ if $$m$$ the multiplicity of $$x-\mu$$ as a factor of $$h$$. Recall that any rational function can be written as $$c\prod_{j=1}^r (x-\mu_j)^{m_j}$$. For a rational function $$h$$ we denote by $$h^+$$ the one with $$h^+(x)=h(x+1)$$. Then clearly $$v(h^+,\mu-1)=v(h,\mu)$$ for all $$h$$, $$\mu$$ and $$v(hr,\mu)=v(h,\mu)+v(r,\mu)$$, $$v(h/r,\mu)=v(h,\mu)-v(r,\mu)$$ for all $$h,r,\mu$$.

We introduce another notation for rational functions $$h$$ and complex $$\mu$$: $$S(h,\mu)=\sum_{k=-\infty}^\infty v(h,\mu+k).$$ Observe that the sum is essentially finite, that is only a finite number of terms is non-zero. We have $$S(h^+,\mu)=S(h,\mu-1)=S(h)$$ for all $$h$$, $$\mu$$ and $$S(hr,\mu)=S(h,\mu)+S(r,\mu)$$, $$S(h/r,\mu)=S(h,\mu)-S(r,\mu)$$ for all $$h,r,\mu$$.

Now we consider two rational functions $$f,g$$ such that $$f(x)f(x+1)=\frac{g(x+1)}{g(x)}.$$ Claim 1: $$S(f,\mu)=0$$ for all $$\mu$$.

Proof: we calculate on the one hand $$S(ff^+,\mu)=S(g^+/g,\mu)=S(g^+,\mu)-S(g,\mu)=0$$, on the other hand $$S(ff^+,\mu)=S(f,\mu)+S(f^+,\mu)=2S(f,\mu)$$.

Claim 2: There exist a rational function $$R(x)$$ and a constant $$c\in\{\pm1\}$$ such that $$f(x)=c\frac{R(x+1)}{R(x)}$$.

$$\renewcommand{\Re}{\mbox{Re}\,}$$ Proof: We consider the finite set $$M$$ of complex numbers $$\mu$$ such that $$0\leq \Re\mu<1$$ and there exists an integer $$m$$ such that $$v(f,\mu+m)\neq0$$. Then $$\beta,\tilde\beta$$ with $$v(f,\beta)\neq0$$ and $$v(f,\tilde\beta)\neq0$$ such that $$\tilde\beta-\beta$$ is an integer are associated to only one element of $$M$$. We write $$f(x)=c\prod_{\mu\in M}\prod_{k=-\infty}^\infty (x-\mu-k)^{m_{\mu,k}}$$ with some constant $$c$$ and integers $$m_{\mu,k}$$. Again, only a finite number of $$m_{\mu,k}$$ is nonzero and, interpreting $$(x-\alpha)^0=1$$ as usual, the product is essentially finite.

Consider now for $$\mu\in M$$ the integers $$n_{\mu,k}=\sum_{\ell=-\infty}^{k-1} m_{\mu,\ell}$$ for $$k\in\mathbb{Z}$$. By Claim 1, we have $$\sum_{k=-\infty}^\infty m_{\mu,k}=0$$ for all $$\mu\in M$$ and therefore $$n_{\mu,k}$$ is non-zero only for a finite number of $$k$$. Therefore $$R(x)=\prod_{\mu\in M}\prod_{k=-\infty}^\infty (x-\mu-k)^{n_{\mu,k}}$$ defines a rational function. By construction, we have $$n_{\mu,k+1}-n_{\mu,k}=m_{\mu,k}$$ for all integers $$k$$. Since $$R(x+1)=\prod_{\mu\in M}\prod_{k=-\infty}^\infty (x+1-\mu-k)^{n_{\mu,k}}= \prod_{\mu\in M}\prod_{k=-\infty}^\infty (x-\mu-k)^{n_{\mu,k+1}},$$ we calculate $$c\,R(x+1)/R(x)=c\prod_{\mu\in M}\prod_{k=-\infty}^\infty (x-\mu-k)^{n_{\mu,k+1}-n_{\mu,k}}= c\prod_{\mu\in M}\prod_{k=-\infty}^\infty (x-\mu-k)^{m_{\mu,k}}=f(x).$$ This proves Claim 2 except for the determination of the constant $$c$$. Observe here, that $$\frac{g(x+1)}{g(x)}$$ can be written as a product of powers $$(x-\mu)^k$$ without a leading constant factor, i.e. leading factor 1. Since $$f(x+1)f(x)=\frac{g(x+1)}{g(x)}$$, the same is true for $$f(x+1)f(x)$$. Now this constant factor is also $$c^2$$. Hence $$c^2=1$$ and therefore $$c\in\{\pm1\}$$.

Claim 3: Any $$R(x)$$ from Claim 2 satisfies $$g(x)=\lambda\,R(x+1)R(x)$$ with some constant $$\lambda$$.

Proof: We calculate $$f(x)f(x+1)=\frac{R(x+2)}{R(x)}=\frac{h(x+1)}{h(x)}$$ where $$h(x)=R(x)R(x+1)$$. Since also $$f(x)f(x+1)=\frac{g(x+1)}{g(x)}$$, the quotient $$q(x)=\frac{g(x)}{h(x)}$$ satisfies $$1=\frac{q(x+1)}{q(x)}$$ or $$q(x+1)=q(x)$$. Since $$q(x)$$ is a rational function, this is only possible if it is a constant, say $$q(x)=\lambda$$. Hence $$g(x)=\lambda R(x)R(x+1)$$. In the complex setting, we can now choose a complex $$\gamma$$ such that $$\gamma^2\lambda=1$$. Replacing $$R(x)$$ by $$\gamma R(x)$$ provides $$R$$ such that $$f(x)=\pm R(x+1)/R(x)$$ and $$g(x)=R(x+1)R(x)$$.

This completes the proof in the complex setting. In the real setting, the essential consideration is to make sure that given real rational $$f,g$$, the construction of the proof of Claim 2 leads to a real rational $$R(x)$$. Now if $$f(x)$$ is real, then $$\mu\in M$$ if and only if the conjugate $$\bar\mu\in M$$ and the multiplicities $$m_{\mu,k}=m_{\bar\mu,k}$$ coincide. The same relation then holds for the multiplicities $$n_{\mu,k}$$ in the construction of $$R(x)$$. This implies that $$R(x)$$ also is a real rational function. Claim 3 also holds in the real setting. Then the only difference to the complex case is that we have to take the sign of $$\lambda$$ into account and therefore can reach either $$g(x)=R(x+1)R(x)$$ or $$g(x)=-R(x+1)R(x)$$.

Second Proof: (Only in the real setting) Consider the rational function $$q(x)=g(x)/f(x)$$. Using the functional equation of $$f,g$$, it satisfies $$\frac{q(x+1)}{q(x)}=\frac{g(x+1)f(x)}{f(x+1)g(x)}=f(x)^2,\ \ \ q(x+1)q(x)=\frac{g(x+1)g(x)}{f(x+1)f(x)}=g(x)^2.$$ If we can show that $$q(x)$$ is essentially a square, that is there exists a rational $$R(x)$$ such that $$q(x)=R(x)^2$$ or $$q(x)=-R(x)^2$$, then we are done.

In order to show that, we use the valuation function $$v$$ introduced in the first proof. For any complex $$\mu$$, we calculate using $$g(x+1)/f(x+1)=f(x)g(x)$$ $$$$\begin{array}{rcl}v(g/f,\mu+1)&=&v({g^+}/{f^+},\mu)=v(fg,\mu)=v(g,\mu)+v(f,\mu)\\&\equiv& v(g,\mu)-v(f,\mu)=v(g/f,\mu)\mod 2.\end{array}$$$$

Therefore all the integers $$v(q,\mu+m)$$, $$m\in\mathbb Z$$, are congruent modulo 2. As these integers are $$0$$ for suffiently large $$m$$, they must all be even. This allows the construction of the wanted $$R(x)$$. Indeed, if $$q(x)=\alpha\prod_{\mu\in\mathcal M}(x-\mu)^{2m_\mu},$$ where $$\alpha$$ is a real number and $$\mathcal M$$ denotes the finite set of $$\mu$$ such that $$v(q,\mu)=2m_\mu\neq 0$$, then we can choose $$\beta$$ such that $$\alpha/\beta^2\in\{\pm1\}$$ and define $$R(x)=\beta\prod_{\mu\in\mathcal M}(x-\mu)^{m_\mu}$$. Since $$q$$ has real coefficients, $$\mu\in\mathcal M$$ if and only if $$\bar\mu\in\mathcal M$$ and we have $$m_\mu=m_{\bar\mu}$$. This implies that $$R(x)$$ also has real coefficients. It satisfies $$q(x)/R(x)^2\in\{\pm1\}$$ and our statement follows.

• This is an aweseome answer and some excelent work. It will take me some time to read and fully understand it. Some coments though : – Harmonic Sun Apr 11 '19 at 23:15
• (1) You conclude that $R$ has to be a real rational function. But if you take $R$ an always imaginary rational function, $f$ and $g$ are both real and solutions of the equation. (2) No need to have a constant $\lambda$ for the general expression of $g$ : if $f(x) = \frac{R(x+1)}{R(x)}$ and $g(x)=R(x)R(x+1)$ with $R$ an arbitrary rational function, the substitution $R=\sqrt{\lambda}R^*$ leads to $f(x) = \frac{R^*(x+1)}{R^*(x)}$ and $g(x)=\lambda R^*(x)R^*(x+1)$ - this is for $\lambda\geq0$, but for $\lambda<0$ one can take take $i\sqrt{\lambda}$. So any solutions of this form can be reached. – Harmonic Sun Apr 11 '19 at 23:15
• @HarmonicSun: You are right. In the complex setting, the solutions can be written $f(x)=\pm R(x+1)/R(x), g(x)=R(x+1)R(x)$ with some rational $R(x)$. In the real setting, I want to reach a real $R$ and then solutions can be written $f(x)=\pm R(x+1)/R(x), g(x)=\pm R(x+1)R(x)$. I will edit my solution. I apologize for the complicated product notation. – Helmut Apr 12 '19 at 8:52
• @HarmonicSun:The idea is simple, though. A vector $v$ of $n$ integers, the sum of which is 0 can be written as $v=(u,0)-(0,u)$, where $u$ is some vector of $n-1$ integers. – Helmut Apr 12 '19 at 8:56
• @HarmonicSun: I have found a second, simpler proof and also included it in my answer together with a few modifications. I am not sure whether I should remove the first proof... – Helmut Apr 13 '19 at 8:45

A fairly broad (sub?)set of solutions is given by the following obvious formula (updated): $$f(x)={R(x+1)\over R(x)},\\[4ex] g(x)={R(x)\cdot R(x+1)}$$ where $$R(x)$$ is an arbitrary rational function.

All previous iterations of this answer seem to have been covered by this formula.

• Nice answer, I did not see it myself, thanks ! – Harmonic Sun Apr 3 '19 at 16:43
• However, this does not include all of the solutions. For instance, for any given $\lambda\in\mathbb{R}^*, (f(x):=1, g(x):=\lambda)$ is a solution that is not included in the set of solutions you gave (exept in the particular case $\lambda=1$) – Harmonic Sun Apr 3 '19 at 21:22
• You've just found an extension to my solution which I overlooked: we may multiply my $g(x)$ by $\lambda$ and still be fine. Whether that will cover all solutions remains unclear, though I think it still won't. – Ivan Neretin Apr 3 '19 at 21:57
• I was right: it won't. – Ivan Neretin Apr 3 '19 at 22:11
• Wait, I've just generalized it a bit more, so the question is now open again. – Ivan Neretin Apr 4 '19 at 13:45