A Fibonacci conjecture: $\sum_{j=1}^n(\sum_{k=1}^jF_k^2)^3=\left(\sum_{j=1}^nF_j\left(\sum_{k=1}^j F_k^2\right)\right)^2$ I have been staring at the identity below for over a year now but I haven't found a way around proving it even though I have tested for $n \le 2000$ using my computer.

$$\sum_{j=1}^{n} \left(\sum_{k=1}^j {F_k}^2 \right)^3 = \left(\sum_{j=1}^{n} F_j \left(\sum_{k=1}^j {F_k}^2 \right)\right)^2 $$

I accidentally discovered it while trying to play with the sum of cubes equalling a square. I am not a mathematician but I have a passion for it so, I think it is evident I don't have enough mathematical techniques to prove a mathematical claim. I also know I should have made a progress on this identity before I seek help but like I said, I have made all the effort I could but I couldn't crack it at all. I suspect the identity looks similar to that of Nicomanchus theorem which states that
$$\sum_{k=1}^{n} k^3 = \left(\sum_{k=1}^{n} k\right)^2 $$
Any help will be appreciated
 A: You could start by known result $F_{1}^{2}+F_{2}^{2}+\dots+F_{n}^{2}=F_{n}F_{n+1}$ (it can be proven simply by induction, see for example  For the Fibonacci numbers, show for all $n$: $F_1^2+F_2^2+\dots+F_n^2=F_nF_{n+1}$). Using this identity in your statement means you want to show:
$$
\sum_{j=1}^n(F_jF_{j+1})^3=\left(\sum_{j=1}^nF_j\left(F_jF_{j+1}\right)\right)^2.
$$
By another known result we have $\sum_{j=1}^nF_j^2 F_{j+1}=\frac{1}{2}F_nF_{n+1}F_{n+2}$, which can be also proven simply by induction. (Note that the induction step amounts to showing $\frac{1}{2}F_nF_{n+1}F_{n+2}+F_{n+1}^2 F_{n+2}=\frac{1}{2}F_{n+1}F_{n+2}F_{n+3}$, which is the same $F_n+2F_{n+1}=F_{n+3}$, which follows quickly from recurrence relation).
So using the second result to simplify the right hand side, your statement is equivalent to
\begin{equation}
\sum_{j=1}^n(F_jF_{j+1})^3=\left(\frac{1}{2}F_nF_{n+1}F_{n+2}\right)^2.\tag{1}
\end{equation}
This seems to be simple enough to attack by induction on its own.
We can verify it is true for $n=1$ thus establishing the base case. For the induction step, assume (1) is true for $n$, we want to show it holds for $n+1$. Write
$$
\sum_{j=1}^{n+1}(F_jF_{j+1})^3=\sum_{j=1}^{n}(F_jF_{j+1})^3+(F_{n+1}F_{n+2})^3=\left(\frac{1}{2}F_nF_{n+1}F_{n+2}\right)^2+(F_{n+1}F_{n+2})^3.
$$
We need to show this is equal $\left(\frac{1}{2}F_{n+1}F_{n+2}F_{n+3}\right)^2$, in other words it remains to show:
$$
\left(\frac{1}{2}F_nF_{n+1}F_{n+2}\right)^2+(F_{n+1}F_{n+2})^3=\left(\frac{1}{2}F_{n+1}F_{n+2}F_{n+3}\right)^2.
$$
Notice both sides have common factor of $F_{n+1}^2F_{n+2}^2$, dividing by it and multiplying by $4$ we are left to show
$$
F_n^2+4F_{n+1}F_{n+2}=F_{n+3}^2\tag{2}.
$$
But this is simple using the recurrence relation few times:
\begin{align}
F_{n+3}^2-F_n^2&=(F_{n+3}-F_n)(F_{n+3}+F_n)\\
&=(F_{n+2}+F_{n+1}-F_n)(F_{n+2}+F_{n+1}+F_n)\\
&=(F_{n+1}+F_n+F_{n+1}-F_n)(F_{n+2}+F_{n+2})\\
&=(2F_{n+1})(2F_{n+2})\\
&=4F_{n+1}F_{n+2}.
\end{align}
Hence the identity required for induction step holds and this completes proof of (1) (and subsequently of the original identity in combination with the previous results).
