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.,
$$ \begin{align} 4 &= F_1^2 + F_1^2 + F_1^2 + F_1^2 \\ &= F_3^2 + F_0^2 + F_0^2 + F_0^2 \end{align} $$

Bruteforce gives that first gap is $24$: the only Fibonacci squares less than $24$ are $0, 1, 4, 9$. We need at least two nines (otherwise sum is at most $9 + 3 \cdot 4 = 21$), so now we have to represent $6$ as sum of two numbers from $0,1,4$. But such sum is either $8$, or at most $5$. Thus $24$ isn't sum of $4$ Fibonacci squares.
It is not hard to find representation for numbers from 17 to 23.
(but of course Empy2's argument is much more elegant then this)