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) $\endgroup$ Mar 9, 2021 at 20:40
  • 2
    $\begingroup$ 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. $\endgroup$ Mar 9, 2021 at 20:41
  • 2
    $\begingroup$ The number has been considered before, of course: A085846 $\endgroup$ Mar 9, 2021 at 21:30

1 Answer 1


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

  • 14
    $\begingroup$ Gelfand has many theorems, but that one is from Gelfond. $\endgroup$
    – abx
    Mar 10, 2021 at 6:15
  • $\begingroup$ Thanks for the correction. I have updated my answer to attribute the theorem to Aleksandr Gelfond (and not Israel Gelfand). $\endgroup$ Mar 10, 2021 at 18:53
  • 3
    $\begingroup$ Nice exercise: show how the above proof breaks down for the similar equation $x^{x+2} = (x+2)^x$, which has solution $x=2$. $\endgroup$ Apr 7, 2021 at 16:32
  • 9
    $\begingroup$ @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". $\endgroup$ Apr 8, 2021 at 3:08
  • 1
    $\begingroup$ @Venkataramana: excellent! $\endgroup$
    – abx
    Apr 8, 2021 at 4:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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