# Fibonacci numbers that are powers?

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 $f_2 = 1$.)

-
If you want to have a formal solution, you could work your way through using fibonacci number's closed form. $f_n= {\phi^n-a^n \over \sqrt{5}}$, and set it equal to $n^k$, and work your way through. (where $\phi$ is the golden ratio and a is the conjugate of $\phi$ – ciceksiz kakarot Feb 10 '13 at 20:41
In your list, $f_{10}$ should be 55, not 44. – John Bentin Feb 11 '13 at 7:59

From Wikipedia: The only nontrivial square Fibonacci number is 144. So there is no $n_4$.

-
More generally, 144 is the largest power (exponent > 1) in the Fibonacci sequence, so $n_k$ does not exist for $k>2$, unless you count powers of 0. – Charles Feb 10 '13 at 20:38
What makes it non-trivial in this context? All squares are, oddly enough, the square of a number. I think that is pretty trivial no matter the square. The nice thing about 144 is that it is not only the square of 12, but it is also the 12th Fibonacci number. Is that trivial? – CogitoErgoCogitoSum Feb 10 '13 at 20:39
If you want to have a formal solution, you could work your way through using fibonacci number's closed form. $f_n= {\phi^n-a^n \over \sqrt{5}}$, and set it equal to $n^k$, and work your way through. (where $\phi$ is the golden ratio and a is the conjugate of $\phi$ – ciceksiz kakarot Feb 10 '13 at 20:39
@vonbrand Theorem 1 in J. H. E. Cohn, "Square Fibonacci Numbers, Etc." Fibonacci Quarterly 2 (1964), 109-113. Check the HTML Version. – azimut Feb 10 '13 at 23:54
According to Wikipedia, the proof is in Bugeaud, Y; Mignotte, M; Siksek, S (2006), "Classical and modular approaches to exponential Diophantine equations. I. Fibonacci and Lucas perfect powers", Ann. Math. 2 (163): 969–1018. – vonbrand Feb 11 '13 at 0:04