Evaluating the generalized continued fraction obtained from the factorization of a bivariate polynomial equation

Question

Happy New Year, MO community!

We need someone expert in Generalized Continued Fractions (GCFs), with a deep knowledge of the GCFs’ convergence properties, to solve the following problem.

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PROBLEM

Given a prime $p>2$, let $x$ and $y$ be real numbers such that $x>y>1$ and $$ x^p-y^p=1 \tag{1}$$ Let us factorize Eq. (1) by $x-y$ as $$ (x-y)^p+pxy(x-y)R=1 \tag{2} $$ where $R$ is a bivariate polynomial $R(x,y) \geq 1$; it is routine to verify that $R=1$ if $p=3$ whereas $R>1$ for any prime exponent $p>3$. Let us rearrange Eq. (2) conveniently as $$ (x-y)((x-y)^{p-2}(x-y)+pxyR)=1 \tag{3} $$ Let us put $a=pxyR$ and $b=(x-y)^{p-2}$ in Eq. (3): $$ (x-y)((b(x-y)+a))=1 \tag{4} $$ with $a>p$ and $b<1$ because, respectively, $xy>1$ and $x-y<1$. From the factorization of Eq. (4) we obtain a recurrence relation: $$ x-y = \frac{1}{a+b(x-y)} \tag{5} $$ with $a>pb$. The recursive Eq. (5) generates a generalized continued fraction (GCF): $$ x-y = \cfrac{1}{a + \cfrac{b}{a + \cfrac{b}{a + \cfrac{b}{a + \ddots \vphantom {cfrac{1}{1}}}}}} \tag{6} $$ Can you prove (or disprove) that the GCF (6) converges to an irrational limit?

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OBSERVATIONS

The irrationality of $x-y$ via the possible convergence of GCF (6) meets only half of the two-fold Lagrange’s criterion recalled in https://mathoverflow.net/q/383280.

PROS: the partial denominator $a$ is at least $p$ times the partial numerator $b$ (they are both real numbers).

CONS: the partial numerator $b$ is smaller than one and the partial denominator $a$ is extremely likely not an integer.

Answer 1

No, $x-y$ does not have to be irrational.

Indeed, take any rational $r>0$ small enough so that $$(1+r)^p<2.$$ Let $$f(y):=(y+r)^p-y^p.$$ Then $f(1+)<1$ and $f(\infty-)=\infty$. Also, $f$ is continuous on $(1,\infty)$. So, $f(y_*)=1$ for some $y_*\in (1,\infty)$. That is, $$x_*^p-y_*^p=1,$$ where $x_*:=y_*+r>y_*>1$, but $x_*-y_*=r$ is rational.