Is it a coincidence that the 12th Fibonacci number is 12 squared?

Question

The first $12$ Fibonacci numbers are: $$\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline n & 0& 1& 2& 3& 4& 5& 6& 7& 8 & 9& 10& 11& 12\\ \hline f_{n} & 0& 1& 1& 2& 3& 5& 8& 13& 21& 34& 55& 89& 144\\ \hline \end{array}$$

When you get to $f_{12}$ you find it is equal to $144$.

What strikes me here is the following:

\begin{align*} f_{12} &= 144\\ 12^2 &=144. \end{align*}

Not only is $f_{12}$ equal to $144$, but so is $12^2$. Here, $n^2= f_n$.

This is the only example as $n^2 > f_n$ is true for all $1<n<12$ (works for $0$ and $1$) and $n^2< f_n$ is true for all $n > 12$.

So is this just a coincidence?

$f_0=0^2$ and $f_1=1^2$ as well. But yes, they are coincidences. — dezdichado, Apr 15 '17 at 19:35
Except from $0,1,144$ the only perfect power in whole sequence is $8$. — kingW3, Apr 15 '17 at 19:41
A small comment: note that because of its growth rate, for any polynomial $p$ there will be only finitely many (perhaps no) $n$ such that $p(n)=f(n)$. — Noah Schweber, Apr 15 '17 at 19:41
@BeastlyGerbil check out the LaTeX align* environment for aligning multiple equations, it is much simpler than setting the spacing manually. — DMcMor, Apr 15 '17 at 19:52
A number $x$ is a Fibonacci-number (contained in the sequence of Fibonacci-numbers), if and only if $5x^2-4$ or $5x^2+4$ (or both) is a perfect square. Maybe this could lead to an explanation ... — Peter, Apr 15 '17 at 20:13
@Peter Here's a link to the question,found it when I was searching for similar questions. — kingW3, Apr 15 '17 at 20:17
@kingW3 Nice, only two perfect powers (if we ignore the trivial ones $0$ and $1$). Would not have expected that and that it was proven! — Peter, Apr 15 '17 at 20:20

Alex · Answer 1 · 2019-03-29T15:41:52.267

This is a very nice manifestation of the Strong Law of Small Numbers formulated by Richard K. Guy: "There aren't enough small numbers to meet the many demands made of them."

A more striking manifestation of the same "law" of small numbers is this: $$ \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline n & 0& 1& 2& 3& 4& 5& 6& 7& 8 & 9& 10& 11& 12\\ \hline \mbox{Fibonacci } f_{n} & 0& 1& 1& 2& 3& 5& 8& 13& 21& 34& 55& 89& 144\\ \hline \hline \mbox{# squares in }[e^{n-1},e^n) & 0& 1& 1& 2& 3& 5& 8& 13& 21& 36& 58& 96& 159\\ \hline \end{array} $$ The last line is OEIS A306486: the number of squares in the interval $[e^{n-1},e^n)$, which happens to be the same as the Fibonacci number $f_n$ for $0\le n\le 8$. Note that both $f_n$ and A306486($n$) grow about as fast as certain geometric progressions; the common ratio is $\phi\approx1.618\ldots$ for the former and $\sqrt{e}\approx1.6487\ldots$ for the latter.

Is it a coincidence that the 12th Fibonacci number is 12 squared?

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