I was asked by an high school student if there is an algebraic way to find the exact value of the solution to the equation \begin{equation}\label{eq} x^{x+1}=(x+1)^x \end{equation} Let us define that with the expression "algebraic way" the student really means "the solution $x$ to the equation is an algebraic number". Now my feeling is that $x$ has to be transcendental but I'm not able to see how to prove it. Note first that there is a unique solution $2<x<3$, more precisely with Wolfram Alpha one can check that $x \approx 2.29...$ I tried to see if I can use a Gelfond-Schneider type of argument but if I write $$x=(x+1)^{\frac{x}{x+1}}$$ then $x$ will be transcendental if I know that $x+1$ is algebraic, but $x+1$ being algebraic is the same as $x$ being algebraic and then I'm stucked. I also tried some Liouville bound on a suitable approximation of $x$ with fractions but I'm not able to control the error coming from the denominators.

One can also take the logarithm of both sides of the equation yielding $$(x+1)\cdot \ln(x)=x \cdot \ln(x+1)$$ I finally tried to use Baker's theorem on linear independence over $\mathbb{Q}$ and $\overline{\mathbb{Q}}$ of logarithms but I'm also stucked because a solution would yield a linear dependence of the logarithm over $\overline{\mathbb{Q}}$ for example, making the theorem impossible to use. Maybe I think that if I'm able to find an expression $$\frac{\ln(x+1)}{\ln(x)}=\ln(f(x,y))$$ where $f(x,y)$ is a solution of a suitable polynomial $P(t) \in \mathbb{Q}[x,y][t]$ then I can use a Baker type argument, but I'm not able to find such an explicit expression for $f(x,y)$.

So I've decided to ask here if some of you have a way to prove or disprove the transcendence of $x$. Thanks in advance.

  • $\begingroup$ Doesn't that follow from Gelfond–Schneider theorem? ($x=x^{\frac{x+1}{x}}$ is transcendental for irrational algebraic $x$ which is impossible) $\endgroup$ – Antoine Labelle Mar 9 at 20:40
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    $\begingroup$ Gelfond-Schneider theorem implies that if x is algebraic, then it's rational. You can show that there are no rational solutions using simple p-adic estimates. $\endgroup$ – Oleg Eroshkin Mar 9 at 20:41
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    $\begingroup$ The number has been considered before, of course: A085846 $\endgroup$ – Jeppe Stig Nielsen Mar 9 at 21:30

The number $x$ is transcendental, and your Gelfond-Schneider argument almost works.

Suppose to the contrary that $x$ is algebraic. Then $x+1$ and $x/(x+1)$ are also algebraic, and so the Gelfond-Schneider theorem guarantees that $x = (x+1)^{\frac{x}{x+1}}$ is transcendental as long as $x/(x+1)$ is irrational.

Claim: $x/(x+1)$ is irrational.

Suppose to the contrary that $x/(x+1)$ is rational and write $x/(x+1) = \frac{a}{b}$ with $a$ and $b$ positive integers with $\gcd(a,b) = 1$. Noting that $x/(x+1) < 1$ forces $b > 1$. One can then rewrite $x = \frac{a}{b-a}$ and $x+1 = \frac{b}{b-a}$. This gives $$ \frac{a}{b-a} = \left(\frac{b}{b-a}\right)^{\frac{a}{b}}. $$ This leads to $a^{b} (b-a)^{a} = b^{a} (b-a)^{b}$. However, since $\gcd(a,b) = 1$, if $p$ is a prime divisor of $b$, then $p$ cannot divide $a$ and $p$ also cannot divide $b-a$ (since if $p | b-a$ and $p | b$, then $p | b - (b-a) = a$). This makes $a^{b} (b-a)^{a} = b^{a} (b-a)^{b}$ impossible since there is a prime number dividing the right hand side that does not divide the left. This is a contradiction. QED Claim

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    $\begingroup$ Gelfand has many theorems, but that one is from Gelfond. $\endgroup$ – abx Mar 10 at 6:15
  • $\begingroup$ Thanks for the correction. I have updated my answer to attribute the theorem to Aleksandr Gelfond (and not Israel Gelfand). $\endgroup$ – Jeremy Rouse Mar 10 at 18:53
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    $\begingroup$ Nice exercise: show how the above proof breaks down for the similar equation $x^{x+2} = (x+2)^x$, which has solution $x=2$. $\endgroup$ – Marc Chamberland Apr 7 at 16:32
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    $\begingroup$ @abx: Cannot help sharing a joke of one of my old teachers. He liked number theory but not Lie theory and used to say that he was " fond of Gelfond but not fand of Gelfand". $\endgroup$ – Venkataramana Apr 8 at 3:08
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    $\begingroup$ @Venkataramana: excellent! $\endgroup$ – abx Apr 8 at 4:40

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