If $ a + b + c \mid a^2 + b^2 + c^2$ then $ a + b + c \mid a^n + b^n + c^n$ for infinitely many $n$

Claim. $a+b+c\mid a^{2^n}+b^{2^n}+c^{2^n}$ for all $n\geq0$.

Proof. By induction: True for $n=0,1$ $\checkmark$. Suppose it's true for $0,\ldots,n$. Note that $$a^{2^{n+1}}+b^{2^{n+1}}+c^{2^{n+1}}=(a^{2^n}+b^{2^n}+c^{2^n})^2-2(a^{2^{n-1}}b^{2^{n-1}}+b^{2^{n-1}}c^{2^{n-1}}+c^{2^{n-1}}a^{2^{n-1}})^2+4a^{2^{n-1}}b^{2^{n-1}}c^{2^{n-1}}(a^{2^{n-1}}+b^{2^{n-1}}+c^{2^{n-1}})$$

and that

$$2(a^{2^{n-1}}b^{2^{n-1}}+b^{2^{n-1}}c^{2^{n-1}}+c^{2^{n-1}}a^{2^{n-1}})=(a^{2^{n-1}}+b^{2^{n-1}}+c^{2^{n-1}})^2-(a^{2^n}+b^{2^n}+c^{2^n})$$

is divisible by $a+b+c$ by the induction hypothesis.

It seems that there's a partial solution.

Suppose that $\mathrm{gcd}(a,a+b+c)=\mathrm{gcd}(b,a+b+c)=\mathrm{gcd}(c,a+b+c)=1$. Then for $n=k\cdot \phi(a+b+c)+1 \, (k=1,2, \ldots )$, where $\phi$ is Euler's function, we have: $$ (a^n+b^n+c^n)-(a^2+b^2+c^2)=a^2 (a^{n-1}-1) + b^2 (b^{n-1}-1) + c^2 (c^{n-1}-1), $$ where all round brackets are divisible by $a+b+c$ according to Euler theorem. Therefore $(a+b+c) \mid (a^n+b^n+c^n)$ for all these $n$.

There's one more solution (it isn't mine). One can even prove that $(a + b + c) \mid (a^n + b^n + c^n)$ for all $n=3k+1$ and $n=3k+2$. It's enough to prove that $a + b + c \mid a^n + b^n + c^n$ => $a + b + c \mid a^{n+3} + b^{n+3} + c^{n+3}$. The proof is here: https://vk.com/doc104505692_416031961?hash=3acf5149ebfb5338b5&dl=47a3df498ea4bf930e (unfortunately, it's in Russian but it's enough to look at the formulae). One point which may need commenting: $(ab+bc+ca)(a^{n-2} + b^{n-2} + c^{n-2})$ is always divisible by $(a+b+c)$ (it's necessary to consider 2 cases: $(a+b+c)$ is odd and $(a+b+c)$ is even).