A recurrence relation for Fibonacci squared The sequence $S_n$ is defined as $$S_1=S_2 =1$$ and for $n\ge 2$, $$S_{n+1}=2(A_n + G_n)$$ where $A_n = \frac {S_n+S_{n-1}}{2}$ is the arithmetic mean and $G_n= \sqrt { S_nS_{n-1} }$ is the geometric mean.
Claim: $$S_n = (F_n)^2$$ where $F_n$ stands for  Fibonacci's numbers.
My solution:
$$S_{n+1}=2(A_n +G_n)= 2(\frac {S_n+S_{n-1}}{2}+\sqrt { S_nS_{n-1} })$$
$$=S_n+S_{n-1}+2\sqrt { S_nS_{n-1} }$$
$$= (
\sqrt { S_n} + \sqrt  S_{n-1} )^2$$
$$\sqrt { S_{n+1}}=\sqrt { S_n} + \sqrt  S_{n-1} \implies $$
$$\sqrt { S_n} =F_n \implies S_n = (F_n)^2$$
 A: $$S_{n+1}=S_n+2\sqrt{S_nS_{n-1}}+S_{n-1}=(\sqrt{S_n}+\sqrt{S_{n-1}})^2$$
So if we let $F_n=\sqrt{S_n}$, then $F_1=F_2=1$ and 
$$F_{n+1}=F_n+F_{n-1}$$
Can you take it from here?
A: You have found "A recurrence relation for Fibonacci squared" (your title) by a certain kind of approach. 
Here is another approach using a third order recurrence relationship for sequence $(F_n^2)$  :
$$F_n^2=2F_{n-1}^2+2F_{n-2}^2-F_{n-3}^2\tag{1}$$
with, of course, initial values 
$$F_1^2=1, \ F_2^2=1, \ F_3^2=4.$$
Proof of (1) :
$$\begin{cases}F_{n}=F_{n-1}+F_{n-2} &\implies&F_{n}^2=(F_{n-1}+F_{n-2})^2\\
F_{n-3}=F_{n-1}-F_{n-2} &\implies&F_{n-3}^2=(F_{n-1}-F_{n-2})^2\end{cases}$$
Adding these relationships gives (1).
(1) can be used in order to find an explicit formula for $F_n^2$. 
Indeed, the characteristic equation associated with  (1) is
$$r^3-2r^2-2r+1=(r+1)(r^2-3r+1)$$
Its roots are 
$$r_1=(-1), \ r_2=\tfrac12(3+\sqrt{5}), \ r_3=\tfrac12(3-\sqrt{5})\tag{2}$$
Taking into account initial conditions, the explicit solution is :
$$F_n^2=-\dfrac25 r_1^n + \dfrac15 r_2^n + \dfrac15 r_3^n \tag{3}$$
Remarks : 
a) (3) could have been as well obtained by squaring 
Binet relationship:
$$F_n=\dfrac{1}{\sqrt{5}}\left(\left(\tfrac{1+\sqrt{5}}{2}\right)^n-\left(\tfrac{1-\sqrt{5}}{2}\right)^n\right).$$ 
b) An interesting generalization of (1) to higher powers of $F_n$ can be found in :  http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/Fibonomials.html#squares
