Wire Hive

$a_{1} = 1$

$a_{2} = 1 + 6$

$a_{3} = 1 + 6 + 6\times 2$

$\vdots$

$a_{n} = 1 + 6\times(1+2+3+\cdots) = 1 + 6\times(\dfrac{n(n-1)}{2}) = 1+3n(n-1) = 3n^2 - 3n +1$

Suppose $a_{n} = m^2$ for some integer $m$ .

Then $a_{n} = 3n^2 - 3n +1 = m^2$ .

$3n^2 - 3n + (1-m^2) = 0$

Solving for $n$ , $n = \dfrac{3 \pm \sqrt{3^2 - 4(3)(1-m^2)}}{2\times 3} = \dfrac{3 \pm \sqrt{3(3 - 4(1-m^2))}}{2\times 3} = \dfrac{3 \pm \sqrt{3(4m^2 - 1)}}{2\times 3} = \dfrac{3 \pm \sqrt{3(2m + 1)(2m - 1)}}{2\times 3}$

Considering the discriminant, $3(2m + 1)(2m - 1)$ is a perfect square, and since $2m + 1$ and $2m - 1$ have absolute difference of $2$ , they are coprime odd numbers. That means, either $2m + 1$ or $2m - 1$ is a perfect square while the other is a multiple of $3$ and another perfect square.

For the latter, the least number is $3 = 3\times 1^2$ while the other number is $1$ , a perfect square. Plugging in these, we will get $n = 1$ and $a_{1} = 1$ , which is our first perfect square.

Then the next possible odd multiple is $27 = 3\times 3^2$ , and then clearly $25 = 5^2$ , where $27-25 = 2$ . Hence, $2m + 1 = 27; m = 13$ .

As a result, the next perfect square is $a_{8} = 13^2 = \boxed{169}$ .