Proof for $\forall n\in \mathbb{N}: 7\mid(1 + 2^{2^n} + 2^{2^{n+1}})$ I am stuck at the following exercise:
Prove that
$\forall n\in \mathbb{N}: 7\mid(1 + 2^{2^n} + 2^{2^{n+1}})$.
I tried to prove the Expression by induction but I cannot find a way to prove the implication
$$7\mid(1 + 2^{2^n} + 2^{2^{n+1}}) \Rightarrow 7\mid(1 + 2^{2^{n+1}} + 2^{2^{(n+1)+1}}).$$
Any help would be appreciated very much.
 A: Suppose $1+2^{2^n}+2^{2^{n+1}}=7k$ and set, for simplicity, $a=2^{2^n}$. Then $2^{2^{n+1}}=a^2$ and $2^{2^{n+2}}=a^4$. Then
\begin{align}
&1+2^{2^n}+2^{2^{n+1}}=1+a+a^2 \\[4px]
&1+2^{2^{n+1}}+2^{2^{n+2}}=1+a^2+a^4
\end{align}
Then
$$
(1+a^2+a^4)-(1+a+a^2)=a^4-a=a(a-1)(a^2+a+1)=7ka(a-1)
$$
so
$$
1+a^2+a^4=7k+7ka(a-1)
$$
A: Let $a_n=2^{2^n}$. We have $a_{n+1}=a_n^2$, hence the sequence $\{a_n\pmod{7}\}_{n\geq 0}$ is periodic from some point on. Let we compile a small table:
$$ \begin{array}{|c|c|c|c|c|c|}\hline n & 0 & 1 & 2 & 3 & \ldots  \\ \hline a_n\pmod{7} & 2 & 4 & 2 & 4 & \ldots\\ \hline\end{array}$$
By induction it is trivial that $a_n\equiv 2\pmod{7}$ if $n$ is even and $a_n\equiv 4\pmod{7}$ if $n$ is odd.
In any case,
$$ 1+a_n+a_{n+1} \equiv 1+2+4 \equiv \color{red}{0}\pmod{7} $$
and that proves the claim.
A: Notice $\ \   \overbrace{x^4\!+x^2\!+1}^{\large f_{\Large n+1}} =\, (x^2\!-x+1)\,\overbrace{ (x^2\!+x+1)}^{\large f_{\Large n}} \,\ $ hence $\ f_n\mid f_{n+1}$
therefore by induction: $\ f_0\mid f_n\ $ for all $\ n \ge 0\,\ $ (in your case $\,f_0 = 7)$
A: Induction will work here.  Assume $1+2^{2^n} + 2^{2^{n+1}}$ is divisible by 7.
Note that $2^{2^{n+2}} = 2^{2^n4} = 16^{2^n} = (14+2)^{2^n} = 14M+2^{2^n}$ for some integer $M$.  Then $1+2^{2^{n+1}} + 2^{2^{n+2}} = 1+2^{2^{n+1}}+(14+2)^{2^n} = 1+2^{2^{n+1}} +2^{2^n}+14M$, which must be a multiple of 7, since the first three terms are from the induction hypothesis.
A: The roots of $x^2+x+1$ over $\mathbb{F}_7$ are $x=2$ and $x=4$.  If $n$ is odd, then $2^n\equiv 2\pmod{3}$, so that $2^{2^n}=2^2=4$ in $\mathbb{F}_7$.  If $n$ is even, then $2^n\equiv 1\pmod{3}$, so that $2^{2^n}=2^1=2$ in $\mathbb{F}_7$.
