Part 1: Moves for $c=-3$: $$1,2,3,4,5$$$$0,2,3,4$$$$0,1,2$$$$-2,2$$$$-3$$Part 2: There can be no integer $n\le -4$. Let the "A-power" of an integer $n$ be defined as the value of the term $a_n$, where $a_0=1$ and $a_n+a_{n+1}=a_{n-2}$ for all integers $n$. Furthermore, let the A-power of a board $C$ be the sum of the A-powers of the numbers written on it. Clearly, the A-power $P_a(C)$ of the board does not change when we perform move (1). Similarly, let the "B-power" of an integer $n$ be defined as the value of the term $b_n$, where $b_0=1$ and $b_n+b_{n+4}=b_{n-1}$ for all integers $n$. Furthermore, let the B-power of a board $C$ be the sum of the B-powers of the numbers written on it. Clearly, the B-power $P_b(C)$ of the board does not change when we perform move (2). The characteristic polynomials of the linear recurrences $a_n$ and $b_n$ are, respectively: $$f_a(x)=x^3+x^2-1$$$$f_a(x)=x^5+x-1$$A quick sketch of $x^3+x^2-1=0$ shows that it has exactly one real root - lets call this root $r$. Since $(x^3+x^2-1)(x^2-x+1)=x^5+x-1$, and $x^2-x+1$ has no real roots, $x^5+x-1=0$ also has exactly one real root - the same root $r$ as $x^3+x^2-1=0$. Thus, $a_n=b_n=r^n$, where $r$ is the single real root of the previously described equations. Then, A-power and B-power are in fact the same thing, so we may just call it "power" as there is no longer any point in differentiating between the two. That is, given any board $C$ with a set of starting numbers, the power of the board, $P(C)=\sum_{i\in (\text{numbers on board})} r^i$, is an invariant. The power after any number of moves will not differ from the power when we have the starting numbers. Then, for all boards $C$, $$P(C)<\sum_{i=1}^{\infty} r^i$$$$\Longrightarrow P(C)<\frac{r}{1-r}$$ Now, suppose there is some number $n\le -4$ on the board. Then, since the power of every integer must be positive (as clearly $0<r<1$), this implies that $P(C)\ge r^{-4}$. However, note that: $$r^5+r-1=0$$$$\Longrightarrow r^5=1-r$$$$\Longrightarrow 1=\frac{r^5}{1-r}$$$$\Longrightarrow r^{-4}=\frac{r}{1-r}$$Then, $P(C)\ge \frac{r}{1-r}$, which directly contradicts our above result. Thus, there can be no integer on the board with value less than or equal to $-4$, so $\boxed{c=-3}$ and we are done.$\blacksquare$