<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] I don't see any way offhand to continue with what you started to finish the proof.
Instead, note that perfect squares have remainders of only $0$, $1$, $4$ or $9$ when divided by $16$. As such, if $a^2$ had a remainder of $1$, the other $2$ would need to have remainders which sum to $15$. This is not possible as the only options are $0$ plus $1$, $4$ or $9$, $1 + 1 = 2$, $1 + 4 = 5$, $1 + 9 = 10$, $4 + 4 = 8$, $4 + 9 = 13$ and $9 + 9 = 18 \equiv 2 \pmod{16}$. Similarly, if $a^2$ had a remainder or $4$ or $9$, the other $2$ also can't add up to give a multiple of $16$.
This means that all of $a^2$, $b^2$ and $c^2$ must have a remainder of $0$ when divided by $16$, i.e., each of $a$, $b$ and $c$ are multiples of $4$. Thus, each of $a^3$, $b^3$ and $c^3$ must be a multiple of $4^3$, i.e., $64$, so their sum is as well.
Update: As Will Jagy's comment indicates, since perfect squares have remainders of only $0$ or $1$ when divided by $4$, this means $a^2 + b^2 + c^2$ can be a perfect square only if $a$, $b$ and $c$ are all even. If you factor out $4$, then for $\left(\frac{a}{2}\right)^2 + \left(\frac{b}{2}\right)^2 + \left(\frac{c}{2}\right)^2$ to be a multiple of $4$ again requires each half value to be even. This means each of $a$, $b$ and $c$ has to be a multiple of $4$ which, as before, results in $a^3 + b^3 + c^3$ being a multiple of $64$.
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