A Double Summation

Let $A$ be the summation above. Note that

$\large \displaystyle 2A + \sum_{i=1}^{100} a_i^{10} = (\sum_{i=1}^{100} a_i^{5})^2$ .

Now, we can see that $a_1^5, a_2^5, ..., a_{100}^5$ are the roots of

$x^{20} - 1$ (Since $(a_i^5)^{20} -1 = a_i^{100} -1 = 0$ )

And that $a_1^{10}, a_2^{10}, ..., a_{100}^{10}$ are the roots of

$x^{10}-1$ (Since $(a_i^{10})^{10} -1 = a_i^{100} -1 = 0$ )

By Vieta's, we can see that

$\sum_{i=1}^{100} a_i^{10} = \sum_{i=1}^{100} a_i^{5} = 0$

This implies that

$\large \displaystyle 2A + \sum_{i=1}^{100} a_i^{10} = 2A + 0 = 0 = (\sum_{i=1}^{100} a_i^{5})^2$

Now, we subtract $0$ from both sides of the equation to get

$2A = 0-0 = 0$

Dividing both sides by 2 gives us

$\frac{2A}{2} = A = \frac{0}{2} = \boxed{0}$

And, thus, our final answer is $\boxed{0}$ .