The Fibonacci sequence is:
$$\left(f_n\right) = \left(0,1,1,2,3,5,8,13,21,34,55,89,144,\dots\right)$$
where we start with $0$ and $1$ and each term in the sequence is the sum of the two previous terms.
Starting the index at $n=0$, I noticed that $f_0=0$, $f_5=5$, and $f_{12}=144$.
Let's just say informally that $0^0=0$ for now. I know this is a controversial issue, but here's a quick argument: the sequence $a_n=0^{1/n}$, converges to $0$ as $n\to\infty$ (and it's obvious that $\frac{1}{n}\to 0$). So to make the pattern clear,
$$f_0=0^0 \qquad f_5=5^1 \qquad f_{12}=12^2$$
There exists an $n_0\in\mathbb{N}$ such that $f_{n_0}=\left(n_0\right)^0$, there exists an $n_1\in\mathbb{N}$ such that $f_{n_1}=\left(n_1\right)^1$, and there exists an $n_2\in\mathbb{N}$ such that $f_{n_2}=\left(n_2\right)^2$.
(1) Does there exist an $n_3\in\mathbb{N}$ such that $f_{n_3}=\left(n_3\right)^3$?
(2) In general, for all $k\in\mathbb{N}$, does there exist some $n_k\in\mathbb{N}$ such that $f_{n_k}=\left(n_k\right)^k$?
(For those of you who don't buy that $0^0=0$, just take $0$ out of $\mathbb{N}$ and start with $f_1 = 1$ and $f_2 = 1$.)
Best Answer
From Wikipedia: The only nontrivial square Fibonacci number is 144. So there is no $n_4$.