$\forall n\in\mathbb N$ prove that at least one of the number $3^{3n}-2^{3n}$,$3^{3n}+2^{3n}$ is divisible by $35$. $3^{3n}-2^{3n}=27^n-8^n=(27-8)(27^{n-1}+27^{n-2}\cdot 8+...+27^1\cdot8^{n-2}+8^{n-1})$
If $n$ is even, 
$3^{3n}+2^{3n}=27^n+8^n=(27+8)(27^{n-1}-27^{n-2}\cdot 8+...-27^1\cdot8^{n-2}+8^{n-1})$
If $n$ is even and a power of $2$, $3^{3n}+2^{3n}$ can't be factorized.
If $n$ is even, $n=m\cdot 2^k,m>1,k>0$ and $m$ is odd $\Rightarrow$
$$27^n+8^n=(27^{2^k}+8^{2^k})\sum_{i=1}^m 27^{(m-i)2^k}(-b^{2^k})^{i-1}$$
How to check divisibility using these cases?
Reference 
 A: if n is odd
$3^{3n} + 2^{3n} = (27+8)(27^{n-1} - 27^{n-2}8 +\cdots + 8^{n-1})$
and $35|(3^{3n} + 2^{3n})$ 
if $n$ is even, $n = 2k:$
$(3^{3n}-2^{3n}) = (3^{n}-2^{n})(3^{2n} + 3^n2^n + 2^{2n})$ 
$(3^{n}-2^{n}) = (3^{2k} - 2^{2k}) = (3^2 - 2^2)(3^{2k-1} +\cdots + 2^{2k-1})\\
5|(3^{2k}-2^{2k})$
What is left?
showing that $7|(3^{6k} - 2^{6k})$
Lets use Fermat's little theorem on this one
when $p$ is prime and $p$ does not divide $a,$ 
$a^{p-1} \equiv 1\pmod p$
$(3^{6k} - 2^{6k}) \equiv 1-1 \equiv 0\pmod 7$
We probably should have brought in the modular arithmetic earlier, but you had already started by factoring.
A: Hint $\ {\rm mod}\ 35\!:\,\ 27\equiv -8\,\ $ so $\,\ \overbrace{27^{\large n}}^{\Large 3^{\Large 3n}}\equiv (-8)^{\large n}\equiv \pm\! \overbrace{8^{\large n}}^{\Large 2^{\Large 3n}}\,$ depending on parity of $n.\  $ QED.
A: HINT.-Modulo $5$ you have $$3^3\equiv 2\pmod5\\2^3\equiv-2\pmod5$$and módulo $7$ you have $$3^3\equiv -1\pmod7\\2^3\equiv1\pmod7$$
You can deduce easily the result for $3^{3n}\pm2^{3n}$
