Is this really an integer solution to $n^{1.01} = n*\ln(n)$

Question

Wolfram alpha says a very large number ($n > 10^{100}$) is an integer solution to $ n^{1.01} = n*\ln(n) $. I'm skeptical and have no way computationally viable way to verify this. Is it true?

Answer 1

Your suspicion turns out to be correct - this equation definitely has no integer solution. To see why this is the case, we note that for any integer $n$, $n^{1.01}$ is an algebraic number, but $\ln(n)$ is a transcendental number for any integer $n>1$m as a corollary to the Lindemann-Weierstrass theorem (indeed, if $\ln(n)\neq 0$ was algebraic, $n=e^{\ln n}$ would be transcendental by the theorem).

Let this be a warning for everyone not to take Wolfram Alpha's word for granted!

Answer 2

Wolfram Alpha has mistook a near-integer for an integer due to its finite precision. I wonder what it would make of $\exp\pi\sqrt{163}$, another famous near-integer. As has been noted, $\ln n$ being transcendental prevents positive integer solutions. However, if we define $n\ln n$ at $n=0$ by continuity, that gives us an uninteresting integer solution.

Answer 3

What Wolfram Alpha provides is one of the solutions, in the real domain, of $$n^{a}=\log(n)\qquad \text{where} \qquad a=\frac 1 {100}$$ which are $$n_1=\left(-\frac{W_0(-a)}{a}\right)^{\frac{1}{a}}\qquad \text{and} \qquad n_2=\left(-\frac{W_{-1}(-a)}{a}\right)^{\frac{1}{a}}$$ For this specific value of $a$, $n_1\approx 2.74602$ and $n_2\approx 1.28544\times 10^{281}$.

Using illimited precision $$n_2-\text{Round}[n_2]\approx -0.327$$