<s>[INST] Prove by induction that $1+2^{2^n}+2^{2^{n+1}}$ is divisible by $7$ As said in the title you have to show by induction that 1+$2^{2^n}+2^{2^{n+1}}$ is divisible by 7. So you start with n=0, that gives 1+2+4=7. So the start is shown.
Let 1+$2^{2^n}+2^{2^{n+1}}$ be divisble by 7 for a n.
I've tried several attempts but I ended up in a mess.
For example the latest attempt:

We have $1+2^{2^{n+1}}+2^{2^{n+2}}$
Adding $2^{2^n}-2^{2^n}$ gives you
$1+2^{2^{n+1}}+2^{2^n}+2^{2^{n+2}}-2^{2^n}$
As the first three summands resemble our Assumption, only $2^{2^{n+2}}-2^{2^n}$ needs to be proved as a multiple of seven.

But I am stuck at trying to show this.
As stated that might be a incorrect approach by myself, so I'm not really sure whether or not that is going in the correct direction. In the end I'd really appreciate some help on this question!
 [/INST] Starting with $n = 0$, we have $1+2^{2^0}+2^{2^{0+1}} = 7$, which is clearly divisible by $7$. Now suppose inductively that $n \ge 1$ and argument holds for all $n$. Then for $n+1$, we have
$$1+2^{2^{n+1}}+2^{2^{n+2}} = 1+2^{2\cdot2^n}+2^{2\cdot2^{n+1}} = 1+(2^{2^n})^2+(2^{2^{n+1}})^2$$
Now and and subtract $(2^{2^n})^2$, we have
$$1+2(2^{2^n})^2+(2^{2^{n+1}})^2-(2^{2^n})^2 = [1+(2^{2^n})^2]^2-(2^{2^n})^2$$ $$ = [1+(2^{2^n})^2+2^{2^n}]\cdot[1+(2^{2^n})^2-2^{2^n}]$$
$$ = [1+2^{2^{n+1}}+2^{2^n}]\cdot[1+2^{2^{n+1}}-2^{2^n}]$$
Here, notice that by inductive hyphothesis, $7|(1+2^{2^{n+1}}+2^{2^n})$. Therefore $7$ divides the whole expression and argument holds for $n+1$. Therefore by induction, argument holds for all $n$.
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