Question

Find the number of 10-tuples $(a_1,a_2,.......,a_{10})$ of integers such that $|a_1|\le 1$ and $a_1^2+a_2^2+a_3^2+........+a_{10}^2-a_1{a}_2-a_2{a}_3-a_3{a}_4-......-a_9{a}_{10}-a_{10}{a}_1=2$.

Answer 1

Let $a_{11}=a_1$.
Multiplying the given equation by 2 we get $(a_1-a_2{)}^2+(a_2-a_3{)}^2+....(a_{10}-a_1{)}^2=4$.
Note that if $a_i-a_{i+1}=\pm 2$ for some i = 1,......, 10, then $a_j=a_{j+1}=0$ for all $j\ne i$ which contradicts the equality $\sum _{i=1}^{10}(a_i-a_{i+1})=0$.
$\therefore a_i=a_{i+1}=1$ for exactly two values of i in {1, 2,......, 10}, $a_i-a_{i+1}=1$ for two other values of i and $a_i-a_{i+1}=0$ for all other values of i.
There are $\left(\begin{array}{c}10\\ 2\end{array}\right)\times \left(\begin{array}{c}8\\ 2\end{array}\right)=45\times 28$ possible ways of choosing these values.
Note that $a_1=1,0$ or 1, so in total there are $3\times 45\times 28$ possible integer solutions to the given equation.