<s>[INST] 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.
 [/INST] A bit longer:

Assume $\tau(n)=\top,n\equiv 1\pmod{3}$
$$1+2^n+2^{2n}=7k\implies 2^{2n}=7k-2^n-1,\;k\in\mathbb Z$$
$\tau(n+1)=\top\implies\;n+1\equiv 2\pmod{3}$
$$1+2^{n+1}+2^{2(n+1)}=1+2\cdot 2^{n}+4\cdot2^{2n}=1+2\cdot2^n+4(7k-2^n-1)\\=28k-2\cdot2^n-3=28k-(2^{n+1}+3)$$
$$7\mid2^{n+1}+3\implies7\mid 2\cdot2^n-4=2(2^n-2)\iff 7\mid2^n-2=2(2^{n-1}-1)\\\iff 7\mid 2^{2n-1}-1$$
$$2^6\equiv 1\pmod{7}\implies 2^{6m}\equiv 1\pmod{7},\;m\in\mathbb N$$
$2n-1=6m\implies 3\mid 2n-1\implies 3\mid 2n+2\implies n\equiv 1\pmod{3}$
</s>