We’re all familiar with the Fibonacci sequence:

$1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89 \ldots$

I was messing around a little on a piece of paper and came to the conclusion that there wouldn’t be any numbers from the sequence that when raised to the power of two, becomes another number contained by the sequence—except that of $1^2$ however, but we’ll leave that out of it.


Although it occurred to me that there probably isn’t any number from the sequence that has this ability other than that of $1$, I don't know how to prove or disprove this.


Whilst I probably could write an application to retrieve the answer for me, I’m more interested in a mathematical solution. Should you clever people prove not to have a great solution, perhaps someone can point me in the right direction?

  • 3
    $\begingroup$ There are very few perfect powers amongst the Fibonacci numbers...in fact, $0,1,8,144$ are the only ones. See this for a (non-elementary) proof. Of course, there may be an easier way to analyze the specific questions you ask. $\endgroup$
    – lulu
    Commented Jan 17, 2017 at 20:02
  • 1
    $\begingroup$ I was going to comment on the trivial $1^2=1$ thing, but you already mentioned it. I like people who think through the trivialities. Nice question! $\endgroup$
    – The Count
    Commented Jan 17, 2017 at 20:15
  • 1
    $\begingroup$ I've tweaked the title a little bit just because the original was an incredibly awkward read - please feel free to change it back if you want. (PS: a fine question!) $\endgroup$ Commented Jan 18, 2017 at 17:54
  • $\begingroup$ @StevenStadnicki Nice work! $\endgroup$
    – D. Ataro
    Commented Jan 18, 2017 at 19:33
  • $\begingroup$ see that at oeis.com $\endgroup$ Commented Dec 14, 2019 at 10:01

4 Answers 4


Let $n>1$. Then $F_n\mid F_m$ if and only if $n\mid m$. (Assuming we start at $F_0=0,F_1=1$.)

This is because we get, inductively, $F_{n+k}\equiv F_{n+1}F_k\pmod{F_n}$, and $F_{n+1}$ and $F_{n}$ are relatively prime.

But you can use Demoivre's formula to show that for $n\geq 2$ that $F_n^2<F_{2n}$.

Specifically, with $\phi=\frac{1+\sqrt{5}}{2}$ and $\rho = \frac{-1}{\phi}=\frac{1-\sqrt{5}}{2}$ we have that:

$$F_{n}=\frac 1{\sqrt{5}}\left(\phi^n-\rho^n\right)$$

And $F_{2n}=F_n\left(\phi^n+\rho^n\right)$. So you need to snow that $\phi^{n}+\rho^n >\frac{1}{\sqrt{5}}(\phi^n-\rho^n)$, or $$(\sqrt{5}-1)\phi^n>-(\sqrt{5}+1)\rho^n$$

  • $\begingroup$ I must say that—although the answer looks very correct, with my current set of skills in mathematics, I’m only able to understand about half of it. Perhaps you could give some more explanations as to why the things you’re writing works. Great work, it looks good—yet—I’m confused! $\endgroup$
    – D. Ataro
    Commented Jan 17, 2017 at 20:17
  • $\begingroup$ Sorry, probably too much to cover in an answer. $\endgroup$ Commented Jan 17, 2017 at 20:18
  • $\begingroup$ @D.Ataro It is worth to read this answer line by line - at least you said you are more interested in a mathematical solution. $\endgroup$ Commented Jan 17, 2017 at 20:20
  • $\begingroup$ Is there perhaps any way you could simplify things a slight bit—or perhaps even write out the final proof somewhat differently? $\endgroup$
    – D. Ataro
    Commented Jan 17, 2017 at 20:21
  • $\begingroup$ It really requires you to tell me how much you know - more than you've told us now, to even begin to start. "Rewrite your answer so I can understand it" isn't much guidance. Help people help you. $\endgroup$ Commented Jan 17, 2017 at 20:23

a non rigorous but maybe easy to understand proof:

Assume such a number $x$ exists, $x > 1, a < x,$ then the sequence goes with, $a, x, x+a, 2x+a, 3x+2a, 5x+3a, ...$

notice the coefficients of the linear expression are in Fibonacci sequence. So that

$ax + (x-a)a,$
$x^2 + a^2 $

are next to each other in the sequence.

But $ax + (x-a)a = x^2 - (x-a)^2 < x^2,$ and the next one $> x^2$. so $x^2$ is not in the sequence.

  • $\begingroup$ This can easily be made rigorous, and I'd encourage you to do so - essentially, if you analyze where your two terms appear in the sequence, it looks like we can bound $F_n^2$ between $F_{2n-1}$ and $F_{2n}$ (and this could be made rigorous using explicit formulas for those values). $\endgroup$ Commented Jan 18, 2017 at 19:41
  • $\begingroup$ @StevenStadnicki, I'm glad you did it. BTW, to get $F_{2n-2}$ and $F_{2n-1}$, it's simple to use the fact $F_{n+k} = F_kF_{n-1} + F_{k+1}F_n$, for $k \ge 1$. $\endgroup$
    – Ryan Y.
    Commented Jan 19, 2017 at 5:30

Working off of Ryan Y's approach:

Note that we have $F_{2n-1} = F_{n-1}^2+F_n^2$; thus, we know that $F_{2n-1}\gt F_n^2$ as long as $F_{n-1}^2\gt 0$, which will happen for all $n\gt 1$.

But we also have $F_{2n-2} = 2F_nF_{n-1}-F_{n-1}^2$, and so $F_n^2\gt F_{2n-2}$ as long as $F_n^2-F_{2n-2}\gt 0$ — or, in other words, $F_n^2-2F_nF_{n-1}+F_{n-1}^2\gt 0$, or $(F_n-F_{n-1})^2\gt 0$. And so this holds whenever $F_n\gt F_{n-1}$, which will hold as soon as $n\gt 2$.

Thus, we have explicitly that $F_{2n-2}\lt F_n^2\lt F_{2n-1}$ for all $n\gt 2$.

(These doubling identities are classical, and there are many ways to derive them; I prefer the matrix-based approach seen in this answer because it's very general and it almost immediately generalizes even further to other linear recurrence sequences.)


This is a trivial proof from a stronger result:

It has been proved that the only squares in the Fibonacci sequence are $1$ and $144.$

Since $12$ is not a Fibonacci number, $1^2=1$ is the only solution, but as you mentioned, we are leaving it out.

See here for more info


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .