Lagrange's Four Square Theorem. Every positive integer $N$ can be written as the sum of four squares, i.e.,
$$N = a^2 + b^2 + c^2 + d^2.$$
Zeckendorf's Theorem. Every integer has a unique representation where it can be written as the sum of non-consecutive Fibonacci numbers. i.e.,
$$N = \sum_i F_{c_i}, c_i \in \mathbb{Z}, c_i \ge 2, c_{i+1} \gt c_i + 1,$$
where $F_k$ is the Fibonacci sequence generated by the recurrence equation $F_{n+2} = F_{n+1} + F_n$ with $F_0 = 0, F_1 = 1$.
Question: Is every positive integer also representable as the sum of four Fibonacci squares?
$$
\begin{align}
0 &= F_0^2 \\
1 &= F_1^2 \\
2 &= F_1^2 + F_1^2 \\
3 &= F_1^2 + F_1^2 + F_1^2 \\
4 &= F_3^2 \\
5 &= F_3^2 + F_1^2 \\
6 &= F_3^2 + F_1^2 + F_1^2 \\
7 &= F_3^2 + F_1^2 + F_1^2 + F_1^2 \\
8 &= F_3^2 + F_3^2 \\
9 &= F_4^2 \\
10 &= F_4^2 + F_1^2 \\
11 &= F_4^2 + F_1^2 + F_1^2 \\
12 &= F_4^2 + F_1^2 + F_1^2 + F_1^2 \\
13 &= F_4^2 + F_3^2 \\
14 &= F_4^2 + F_3^2 + F_1^2 \\
15 &= F_4^2 + F_3^2 + F_1^2 + F_1^2 \\
16 &= F_3^2 + F_3^2 + F_3^2 + F_3^2
\end{align}
$$
Note that in the sums above, if we have fewer than $4$ squares, we can always add $F_0^2 = 0^2$ to make it a four square representation.
The sum of four Fibonacci squares representation, if it exists, is not necessarily unique for an integer. e.g.,
$$
\begin{align}
4 &= F_1^2 + F_1^2 + F_1^2 + F_1^2 \\
&= F_3^2 + F_0^2 + F_0^2 + F_0^2
\end{align}
$$
Best Answer
If $N$ is big, there are only about $\ln N$ Fibonacci squares below $N$. There are then at most $(\ln N)^4$ different sums below $N$. That count eventually grows more slowly than $N$, so gaps will appear.
I don't know when the first gap will be