10 Tuples

The desired condition is that $(a_1 - a_2)^2 + (a_2 - a_3)^2 + (a_3 - a_4)^2 + \cdots + (a_9 - a_{10})^2 + (a_{10} - a_1)^2 \; = \; 4$ and so the integers $b_1= a_1 - a_2$ , $b_2 = a_2 - a_3$ , ..., $b_9 = a_9 - a_{10}$ , $b_{10} = a_{10} - a_1$ are such that $-2 \le b_j \le 2$ for all $j$ , while $b_1 + b_2 + \cdots + b_{10} = 0 \hspace{2cm} b_1^2 +b_2^2 + \cdots + b_{10}^2 = 4$ We cannot have a single $b_j$ equal to $\pm2$ , with the rest zero, and so we deduce that two of the $b_j$ must be equal to $1$ , and another two must be equal to $-1$ . There are $\binom{10}{4}$ ways of choosing which of the $10$ numbers $b_j$ are to be nonzero.

If we consider the four nonzero numbers:

• if they appear in any of the orders $1,1,-1-1$ or $1,-1,-1,1$ or $-1,-1,1,1$ or $-1,1,1,-1$ , then the values of $a_j$ increase and decrease by $2$ over the length of the sequence. Thus there is only one possible pattern of $a_j$ which yields each of these patterns of nonzero $b_j$ .

• if they appear in either of the orders $1,-1,1,-1$ or $-1,1,-1,1$ , then the values of $a_j$ only need to be able to rise and fall by $1$ over the sequence. Thus there are two possible $a_j$ sequences for each of these $b_j$ patterns.

Thus we have a total of $\binom{10}{4} \times (4 \times 1 + 2 \times 2) = 1680$ sequences of $a_j$ , making the answer $\boxed{3780}$ .