<s>[INST] If $a^2 + b^2 + c^2$ is divisible by $16$, then show that$ a^3 + b^3 + c^3$ is divisible by $64$. Where $a, b, c \in \mathbb{Z}$. If $a^2 + b^2 + c^2$ is divisible by $16$, then show that $a^3 + b^3 + c^3$ is divisible by $64$; where $a, b, c \in \mathbb{Z}$.
I began by proving that if $(a^3+b^3+c^3) -(a^2+b^2+c^2)$ is divisible $16$, then $a^3+b^3+c^3$ will be divisible by $64$. That is equal to $a^3-a^2+b^3-b^2+c^3-c^2$ $= a(a^2-a)+b(b^2-b)+c(c^2-c)$.
But this doesn't give a clear path for the question.
Any help? 
 [/INST] We may write:
$$16\big|a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ac)$$
⇒ $$4\big| a+b+c$$
Also:
$$8\big| ab+bc+ac$$
Since the phrases are symmetric if one is a multiple of 4 others will also be, Hence a, b and c are multiple of 4 so the sum of their cubes are  a multiple $4^3=64$.
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