Powers for the sum of three squares

Question

Say $n$ can be represented as a sum of three non-zero squares. (i.e. $n = a^2 +b^2+ c^2$, for some $n,a,b,c \in \mathbb{N}$)

Is it possible that every natural power of $n$ is also a sum of three non-zero squares? (i.e. $n^k = x^2+y^2+z^2$ for $x,y,z,k \in \mathbb{N}$)

Answer 1

Yes, more generally, if, $$N = x_1^2+x_2^2+\dots+x_m^2$$ then one can always find integer $y_i$ such that, $$N^k = y_1^2+y_2^2+\dots+y_m^2$$

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Proof: Use the expansion of

$$(a+i\sqrt{w})^k$$

To get,

$$a^2 + w = N,\\ (a^2 - w)^2 + (2 a)^2 w = (a^2 + w)^2,\\ (a^3 - 3 a w)^2 + (3 a^2 - w)^2 w = (a^2 + w)^3,\\ (a^4 - 6 a^2 w + w^2)^2 + (4 a^3 - 4 a w)^2 w = (a^2 + w)^4$$

and so on, then assume $w$ to be a sum of squares $w=b^2+c^2+\dots$

For example, for $k=3$,

$$\small\big(a^3 - 3 a (b^2+c^2+\dots)\big)^2 + \color{blue}{\big(3 a^2 - (b^2+c^2+\dots)\big)^2 \big(b^2+c^2+\dots\big)} = (a^2 + b^2+c^2+\dots)^3$$

then distribute the blue term. I trust the pattern for other $k$ is easy to discern.

Answer 2

Yes, this is possible. Take $n=3$. Clearly $3=1^2+1^2+1^2$. Now any power of $3$ is also the sum of three squares by the Three-Squares Theorem, because $3^k$ is not of the form $4^n(8m+7)$. This may include zero summands, though. Otherwise one has to give a direct solution. If $n=3^k$ with $k=2m+1$, then $3^k=(3^m)^2+(3^m)^2+(3^m)^2$. If $n=3^k$ with $k=2m$, then $$ 3^k=3^{2m}=(3^{m-1})^2+(2\cdot 3^{m-1})^2+(2\cdot 3^{m-1})^2 $$