10. Let $N_{4}(n)$ denote the total number of solutions to $n=x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}$ (different orders and signs are considered as different solutions), and let $\sigma(n)$ denote the sum of the positive divisors of $n$, i.e., $\sigma(n) = \sum_{d \mid n} d$ (for example, $\sigma(6) = 1 + 2 + 3 + 6 = 12$). This problem aims to use the discussion and conclusions about the total number of solutions $N(n)$ to $x^{2} + y^{2} = n$ to prove:
$$N_{4}(n)=\left\{\begin{array}{ll} 8 \sigma(n), & 2 \nmid n, \\ 24 \sigma(m), & n=2^{k} m, k \geqslant 1, 2 \nmid m . \end{array}\right.$$
(i) Suppose $2 \nmid n$, and let $M(n)$ be the total number of solutions to the indeterminate equation $4 n = u_{1}^{2} + u_{2}^{2} + u_{3}^{2} + u_{4}^{2}$, where $2 \nmid u_{j} (1 \leqslant j \leqslant 4)$. Prove that $M(n) = \sigma(n)$.
(ii) Suppose $2 \nmid n$. Prove that $N_{4}(2 n) = 3 N_{4}(n)$, $N_{4}(2 n) = N_{4}(4 n)$, and $N_{4}(4 n) = 16 \sigma(n) + N_{4}(n)$.
(iii) Derive the desired conclusion from (i) and (ii).
(iv) Verify the correctness of the above conclusions for $n = 105, 210$ (Hint: To prove (i), use
$$\begin{aligned} M(n) & =\sum_{a+b=2 n} \frac{N(2 a)}{4} \frac{N(2 b)}{4} \\ & =\sum_{a+b=2 n}\left(\sum_{s \mid 2 a} h(s)\right)\left(\sum_{t \mid 2 b} h(t)\right) \\ & =\sum_{a x+y=2 n} h(s t) \end{aligned}$$

Here the variables $s, t, x, y$ take positive odd values. Divide the final sum into two parts: $s = t$ and $s \neq t$. The sum of the part where $s = t$ equals $\sigma(n)$, and the other part is zero.