Very Complex Sequence

$\begin{aligned} 2a_n^2+51a_{n-1}^2+25a_{n-2}^2-20a_na_{n-1}-10a_{n-1}a_{n-2}&=0 \\ 2(a_n^2-10a_na_{n-1}+25a_{n-1}^2)+(a_{n-1}^2-10a_{n-1}a_{n-2}+25a_{n-2}^2)&=0 \\ 2(a_n-5a_{n-1})^2+(a_{n-1}-5a_{n-2})^2&=0 \end{aligned}$

As the sum of square of real numbers is $0$ , we can conclude that $a_n-5a_{n-1}=a_{n-1}-5a_{n-2}=0$ , which means $a_n=5a_{n-1}$ . It is obvious that it is a geometric progression. As $a_1=2$ , $a_2=10$ and $a_{2018}=2\times 5^{2017}$ .

Hence, the number of positive divisor of $a_{2018}$ is $(1+1)\times (2017+1)=\large{4036}$ .