Show that $1+2^n+2^{2n}$ is divisible by $7$, when $n$ is not a multiple of $3$ Problem taken from a paper on mathematical induction by Gerardo Con Diaz. Although it doesn't look like anything special, I have spent a considerable amount of time trying to crack this, with no luck.
Show that $1+2^n+2^{2n}$ is divisible by $7$, when $n$ is not a multiple of 3.

Whenever a number is not divisible by 3, it leaves remainder as 1 or 2. So, any number which is not divisible by 3 is of the form $n=3l+1$ or $n=3l+2$.
Now, when $n=3l+1$, we have $1+2^{3l+1}+2^{2(3l+1)}=1+2+4$ mod $7$.
and when 
$n=3l+2$, we have $1+2^{3l+2}+2^{2(3l+2)}=1+4+16$ mod $7$ and $1+4+16=0$ mod $7$.
