# How to prove that the solution to $x^{x+1}=(x+1)^{x}$ is transcendental?

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 $$$$\label{eq} x^{x+1}=(x+1)^x$$$$ Let us define that with the expression "algebraic way" the student really means "the solution $$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, 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.

• Doesn't that follow from Gelfond–Schneider theorem? ($x=x^{\frac{x+1}{x}}$ is transcendental for irrational algebraic $x$ which is impossible) Mar 9, 2021 at 20:40
• 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. Mar 9, 2021 at 20:41
• The number has been considered before, of course: A085846 Mar 9, 2021 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

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