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.
Best Answer
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