Is there a Fibonacci number whose square is another Fibonacci? Introduction
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.
Research
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.
Moreover
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?
 A: 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 $$(\sqrt{5}-1)\phi^n>-(\sqrt{5}+1)\rho^n$$
A: 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.
A: 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.)
A: 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
