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The case $ n=1$ yields trivially $ x_1=a$ (only equation in the system). Assume $ n\geq4$. Then, $ a^4=(x_1^2+...+x_n^2)^2=x_1^4+...+x_n^2$, yielding $ \sum x_i^2x_j^2=0$ for $ 1\leq i<j\leq n$, or at most one of the $ x_i$ is nonzero, and it must equal $ a$. If $ n=3$, we find $ a^2=x_1^2+x_2^2+x_3^2=(x_1+x_2+x_3)^2$ yielding $ x_1x_2+x_2x_3+x_3x_1=0$, or using that $ x_1^3+x_2^3+x_3^3-x_1x_2x_3=(x_1+x_2+x_3)(x_1^2+x_2^2+x_3^2-x_1x_2-x_2x_3-x_3x_1)$, we conclude that $ a^3-x_1x_2x_3=a(a^2-0)$, or $ x_1x_2x_3=0$, and wlog at least $ x_3=0$. But then $ x_1x_2=0$, and again at most one of the $ x_i$ is nonzero, and it must equal $ a$. Finally, if $ n=2$, then $ a^2=x_1^2+x_2^2=(x_1+x_2)^2$, and $ x_1x_2=0$, and again at most one of the $ x_i$ is nonzero, and it must equal $ a$.

$S_k=x_1^k+x_2^k+...x_n^k$ and let $\alpha_k, k=0,1,...,n$ denotes the $k$th elementary symmetric polynomial in $x_1,x_2,...x_n$ Using Newton's formula, $k\alpha_k=S_1\alpha_{k-1}-S_2\alpha_{k-2}+......+(-1)^kS_{k-1}\alpha_1+(-1)^{k-1}S_k$ This equation has a root for $\alpha_1=a,\alpha_k=0$ so by vieta $x_1,x_2,...x_n$ are the roots of the polynomial $x^n-ax^{n-1}$. So the roots are permutation of $a, 0,0,...,0$