Is the permutations of the solutions considered different?

First note that we can replace $x_i$ with $-x_i$ to get a different set of solutions if $x_i \neq 0$. Since $m>0$ there always exists $x_i \neq 0$. So for odd $n$ we get that the number of solutions is even. Now we show $n$ satisfies conditions $\Rightarrow 2n$ satisfies conditions. Let $n=2^a \cdot q$ where $q$ odd. Let $f(n,m) = $ number of integer solutions to $x_1^2 + x_2^2 + \dots + x_n^2 = m$ Consider $\sum_{i=1}^{n} x_i^2 = b, \sum_{i=n+1}^{2n} x_i^2 = m-b$ for $1 \leq b \leq m-1$ $2^{2(a+1)}$ divides number of integer solutions for each pair of equations for each $b$, hence $2^{a+2}$ divides number of integer solutions as well. As for $\sum_{i=1}^{n} x_i^2 = 0, \sum_{i=n+1}^{2n} x_i^2 = m$ and $\sum_{i=1}^{n} x_i^2 = m, \sum_{i=n+1}^{2n} x_i^2 = 0$ The total number of integer solutions to above two pairs of equations is $2f(n,m)$ which is divisible by $2^{a+2}$. So $f(2n,m)=$ sum of number of solutions to all above pairs of equations, which is divisible by $2^{a+2}$. So from $n$ odd works we extend to all $n$ works.