Find all the values of $n \in N$ such that $n^2 = 2^n$.
Problem
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Tags: number theory, diophantine, Diophantine equation
MathNerd555
30.09.2021 23:22
If I recall correctly the only solutions among the natural numbers are 2 and 4? I might be wrong; I just know the answers since I watched a video on it
jasperE3
30.09.2021 23:24
It's actually not hard to find all $n\in\mathbb R$ with calculus (or more intelligent bounding).
Jwenslawski
30.09.2021 23:31
I think you could use the Lambert W function.
jasperE3
30.09.2021 23:33
Jwenslawski wrote: I think you could use the Lambert W function. I believe that there is no efficient way to use the Lambert W function. Please post a solution with Lambert W if you have one.
OlympusHero
01.10.2021 00:02
I claim that for all $n>4$ we have $n^2<2^n$. We'll show this by induction. The base case $n=5$ clearly holds true, so it suffices to show $\left(\frac{n+1}{n}\right)^2<2$. But for all $x>n$, we have $\frac{x+1}{x} < \frac{n+1}{n}$, so it suffices to show this holds true for $n=5$, and it does. Clearly $n^2=2^n$ cannot hold true for negative values of $n$, so we must only test $n=0,1,2,3,4$. Out of these, only $n=\boxed{2,4}$ work.