Let $a_k$ be the number of ordered 10-tuples $(x_1, x_2,\ldots,x_{10})$ of nonnegative integers such that $x_1^2 + x_2^2 +\cdots+ x _{10}^2 = k.$ Let $$ b_k= \begin{cases} 0 & \text{if}\,\, a_k\,\, \text{is even}\\ 1 & \text{if}\,\, a_k\,\,\text{is odd} \end{cases} $$
Find $\sum_\limits{i=1}^{2012}b_{4i}$.
Any help is appreciated.
For those curious, the solution which I have trouble comprehending is item 1 from the CHMMC-2012 Spring Tiebreakers.
I'm having trouble understanding the first part of the solution:
...We can pair each 10-tuple with its reverse without changing the parity of the number of solutions. Only those 10-tuples which are palindromes are unpaired, so $x_i = x_{11−i}$ and so we can consider the following equation without changing the $b_k$ sequence: $2x_1^2 + 2x_2^2 + 2x_3^2 + 2x_4^2 + 2x_5^2 = k$
I know this is a basic question, but why can you assume $x_i = x_{11−i}$ and how does it not change parity of number of solutuins to the equation?