It's easy to check that the equation is equivalent to $\sum(a_i-a_{i+1})^2 = 4$ (indices $\pmod{10}$ of course). Now let the differences be the numbers $a_i - a_{i+1}$. So either one of the differences is $\pm{2}$ (Case 1), or 4 of the differences are $\pm{1}$ (Case 2) Case 1 is impossible since the differences sum to 0. In case 2, 2 of the differences are $1$, 2 of the differences are $-1$, since the differences sum to 0. Now for this we have another 2 cases. There are $\binom{10}{4} = 210$ ways to make some of the differences $\pm{1}$. Now there are 2 ways to make those differences either $1$, or $-1$, so that no 2 consecutive differences which are either $1$ or $-1$ are the same. For these sequences, we can get 2 different sequences, one of only 0's and 1's or one of only 0's, -1's. So we get $2 \cdot 2 = 4$ in this case. For the other 4 ways, we get either 2 1's consecutive, or -1's consecutive, and we get a unique sequence this way, since we must use up to 1, then decrease to -1 or vice versa. So $4$ ways. So the answer is $(4+4) \cdot 210 = 1680$.

That's out by $2000$.

Answer $3780$ As $\sum_{r=1}^{10} a_r-a_{r+1}=0$ where $a_{11}=a_1$, Out of $10$ squares Two will square of $1$ Two will be square of $-1$ Rest $0$ and $a_1$ can take $3$ values $\pm1,0$ Hence sequences $= 3\times \binom{10}{4}\times \binom{4}{2}=3780$

Whats the correct answer?

3780 is the correct answer #4 is the correct explanation