existence of solutions of $a^n+b^n+c^n=6^n$ $$2^2+4^2+4^2=6^2$$
$$3^3+4^3+5^3=6^3$$ 
There is no such thing for exponent 4. It's not hard to prove that the only integer solutions of $a^4+b^4+c^4=6^4$ are trivial ones.
For which $n$ there exist $a,b,c\in\mathbb{N}$ such that $a^n+b^n+c^n=6^n$?
(own question, a very probable answer is: $2$ and $3$ only).
 A: For $n \geq 7,$
$$ \left( \frac{5}{6} \right)^n < \frac{1}{3}  $$
A: There are only solutions for $n=2,3$ as you suspect.
I assume you're looking for non-trivial solutions, so $0\leq a,b,c \leq 5$. Assuming that $a,b,c$ satisfy your equation, then
$$6^{n}=a^{n}+b^{n}+c^{n}
\leq 5^{n}+5^{n}+5^{n} = 3\times 5^{n} \\
\Rightarrow n \leq \frac{\log(3)}{\log(6)-\log(5)}\approx 6.0257$$
So that leaves us with the two cases $n=5$ and $n=6$. For $n=6$, Fermat's little theorem implies that $a^{6}\equiv 1 (\textrm{mod}\,7)$ if $1\leq a \leq 5$ and $a^{6}\equiv 0 (\textrm{mod}\,7)$ if $a=0$. As $6^{6}\equiv 1 (\textrm{mod}\,7)$, we must have that two of $a,b,c$ are zero, which leaves us with no solution.
Finally for $n=5$, first observe that at least one of $a,b,c$ must be $5$. Otherwise 
$$6^{5}=a^{5}+b^{5}+c^{5}
\leq 3\times 4^{5}$$
which can be checked to be false. So
$$6^{5}-5^{5}=a^{5}+b^{5}$$
Now we just have to plug in the numbers $0$ to $5$ for $a$ and check if the result can be satisfied for $b$. It is easily seen that there are no such solutions.
