nicetry007 wrote: We will show that $ a+b+c$ divides $ a^{2^{k}}+\; b^{2^{k}}+\; c^{2^{k}}$ for $ k \in \{0\;,\; 1\;,\; 2\; \cdots \;\}$. It is given in the problem that the result holds for $ k \;= \;0$ and $ k \;= \;1$. We will extend it to every positive integer by induction on $ k$. Suppose $ a+b+c$ divides $ a^{2^{n-1}}+\; b^{2^{n-1}}+\; c^{2^{n-1}}$ and $ a^{2^{n}}+\;b^{2^{n}}+\;c^{2^{n}}$. We then have $ a+b+c \; |\;\left(a^{2^{n-1}}+\; b^{2^{n-1}}+\; c^{2^{n-1}}\right)^{2}-\left(a^{2^{n}}+\;b^{2^{n}}+\;c^{2^{n}}\right)\; = \; 2\left(a^{2^{n-1}}b^{2^{n-1}}+b^{2^{n-1}}c^{2^{n-1}}+c^{2^{n-1}}a^{2^{n-1}}\right)$. $ 2\;\left(a^{2^{n-1}}b^{2^{n-1}}+b^{2^{n-1}}c^{2^{n-1}}+c^{2^{n-1}}a^{2^{n-1}}\right)^{2}$ $ = 2\;\left(a^{2^{n}}b^{2^{n}}+\; b^{2^{n}}c^{2^{n}}+\; c^{2^{n}}a^{2^{n}}+\;2\;\left(a^{2^{n-1}}b^{2^{n}}c^{2^{n-1}}+\;a^{2^{n-1}}b^{2^{n-1}}c^{2^{n}}+\; a^{2^{n}}b^{2^{n-1}}c^{2^{n-1}}\;\right)\; \right)$ $ = 2\;\left(a^{2^{n}}b^{2^{n}}+\; b^{2^{n}}c^{2^{n}}+\; c^{2^{n}}a^{2^{n}}+\;2\;a^{2^{n-1}}b^{2^{n-1}}c^{2^{n-1}}\;\left(a^{2^{n-1}}+\;b^{2^{n-1}}+\; c^{2^{n-1}}\;\right)\; \right)$ Since $ a+b+c \; |\;2\left(a^{2^{n-1}}b^{2^{n-1}}+b^{2^{n-1}}c^{2^{n-1}}+c^{2^{n-1}}a^{2^{n-1}}\right)\;$, we have $ a+b+c \; |\;2\left(a^{2^{n-1}}b^{2^{n-1}}+b^{2^{n-1}}c^{2^{n-1}}+c^{2^{n-1}}a^{2^{n-1}}\right)^{2}$ $ \Rightarrow a+b+c \; |\;2\;\left(a^{2^{n}}b^{2^{n}}+\; b^{2^{n}}c^{2^{n}}+\; c^{2^{n}}a^{2^{n}}+\;2\;a^{2^{n-1}}b^{2^{n-1}}c^{2^{n-1}}\;\left(a^{2^{n-1}}+\;b^{2^{n-1}}+\; c^{2^{n-1}}\;\right)\; \right)$ $ \Rightarrow a+b+c \; |\;2\;\left(a^{2^{n}}b^{2^{n}}+\; b^{2^{n}}c^{2^{n}}+\; c^{2^{n}}a^{2^{n}}\; \right)\;$ (as $ a+b+c \;|\; a^{2^{n-1}}+\; b^{2^{n-1}}+\; c^{2^{n-1}}$ ) $ \Rightarrow a+b+c \; |\;\left(a^{2^{n}}+\; b^{2^{n}}+\; c^{2^{n}}\right)^{2}-2\;\left(a^{2^{n}}b^{2^{n}}+\; b^{2^{n}}c^{2^{n}}+\; c^{2^{n}}a^{2^{n}}\; \right)\;$ ( as $ a+b+c \;|\; a^{2^{n}}+\; b^{2^{n}}+\; c^{2^{n}}$ ) (i.e.) $ a+b+c \; |\;a^{2^{n+1}}+\; b^{2^{n+1}}+\; c^{2^{n+1}}$. Hence, $ a+b+c \; |\;a^{2^{k}}+\; b^{2^{k}}+\; c^{2^{k}}$ for all $ k \in \{0\;,\; 1\;,\; 2\; \cdots \;\}$.

Just use the identity $ 2((x + y)^4 + x^4 + y^4) = ((x + y)^2 + x^2 + y^2)^2$. So if $ x + y + z \equiv 0 \equiv x^2 + y^2 + z^2 \pmod{a + b + c}$ it follows from the identity that: $ x^4 + y^4 + z^4 \equiv (y + z)^4 + y^4 + z^4 \equiv (y + z)^2 + y^2 + z^2 \equiv \\ x^2 + y^2 + z^2 \equiv 0 \pmod{a+b+c}$ Continuing with induction, if $ a^{2^k} + b^{2^k} + c^{2^k} \equiv 0 \pmod{a + b + c}$ for all $ k \leq n$ then: $ a^{2^{n+1}} + b^{2^{n+1}} + c^{2^{n + 1}} \equiv (b^{2^{n - 1}} + c^{2^{n-1}})^4 + (b^{2^{n - 1}})^4 + (c^{2^{n-1}})^4 \equiv \\ (b^{2^{n - 1}} + c^{2^{n - 1}})^2 + (b^{2^{n - 1}})^2 + (c^{2^{n - 1}})^2 \equiv a^{2^n} + b^{2^n} + c^{2^n} \equiv 0 \pmod{a+b+c}$

WLOG we can assume that $(a,b,c)=1$ then if $(a,a+b+c)=(b,a+b+c)=(c,a+b+c)=1$ we can easily see that for $\forall t\in N: n=t\times \varphi(a+b+c)+2$ has the property because $a^{t\times \varphi(a+b+c)+2}+b^{t\times \varphi(a+b+c)+2}+c^{t\times \varphi(a+b+c)+2} \equiv a^2+b^2+c^2 \equiv 0 \mod (a+b+c)$ now we prove that the condition above always happens. let $a+b+c=\Pi p_i^{\alpha_i}$ if $p_i$ is a prime factor of one of $a,b,c$, for example $a$ then because $(a,b,c)=1$ we get $p_i \nmid b,c$ and $p_i \mid b+c, p_i\mid b^2+c^2 \implies p_i \mid 2bc \implies p_i=2$ now because $a+b+c$ is even it means one of them is even and two others must be odd note that three of them cannot be even because $(a,b,c)=1$ $\implies a^2+b^2+c^2 \equiv 6$ or $2 \mod 8$ which implies $2 \mid a+b+c, 4\nmid a+b+c$ and then because $2 \mid a^{t\times \varphi(a+b+c)+2}+b^{t\times \varphi(a+b+c)+2}+c^{t\times \varphi(a+b+c)+2}$ and $a^{t\times \varphi(a+b+c)+2}+b^{t\times \varphi(a+b+c)+2}+c^{t\times \varphi(a+b+c)+2} \equiv a^2+b^2+c^2 \equiv 0 \mod p_i^{\alpha_i}$ implies that $a+b+c \mid a^{t\times \varphi(a+b+c)+2}+b^{t\times \varphi(a+b+c)+2}+c^{t\times \varphi(a+b+c)+2}$ and we're done!