<s>[INST] Divisibility by 7 Proof by Induction Prove by Induction that
$$7|4^{2^{n}}+2^{2^{n}}+1,\; \forall n\in \mathbb{N}$$
Base case:
$$
\begin{aligned}
7&|4^{2^{1}}+2^{2^{1}}+1,\\
7&|7\cdot 3
\end{aligned}$$ Which is true.
Now, having $n=k$, we assume that:
$$7|4^{2^{k}}+2^{2^{k}}+1,\;\; \forall k\in \mathbb{N}$$
We have to prove that for $n=k+1$ that,
$$7|4^{2^{k+1}}+2^{2^{k+1}}+1,\;\; \forall k\in \mathbb{N}$$
We know that if $a$ is divisible by 7 then $b$ is divisible by 7 iff $b-a$ is divisible by 7.
Then,
$$
\begin{aligned}
b-a &= 4^{2^{k+1}}+2^{2^{k+1}}+1 - (4^{2^{k}}+2^{2^{k}}+1)\\
&= 4^{2^{k+1}}+2^{2^{k+1}} - 4^{2^{k}}-2^{2^{k}}\\
&= 4^{2\cdot  2^{k}}+2^{2\cdot  2^{k}} - 4^{2^{k}}-2^{2^{k}}
\end{aligned}
$$
I get stuck here, please help me.
 [/INST] Divisibility by $7$ is congruence to zero modulo $7.$
So we might get some insights by looking at the numbers' congruences mod $7$.
Note that $x^{2^{k+1}} = x^{2^k\cdot 2} = \left(x^{2^k}\right)^2.$
That is, every time we add $1$ to the exponent $k$ in
$2^{2^k}$ or $4^{2^k}$, we square the number.
So let's try applying these two ideas: do the arithmetic modulo $7$; and try the first few values of $n$ and see what happens.
So what happens is
\begin{align}
2^2 \equiv 4 \pmod7,\\
4^2 \equiv 2 \pmod7.\\
\end{align}
That is,
\begin{align}
\text{for } n &= 1,
& 4^{2^1} + 2^{2^1} + 1 = 4^2 + 2^2 + 1 &\equiv 2 + 4 + 1 \pmod7,\\
\text{for } n &= 2,
& 4^{2^2} + 2^{2^2} + 1 \equiv 2^2 + 4^2 + 1 &\equiv 4 + 2 + 1 \pmod7,\\
\text{for } n &= 3,
& 4^{2^3} + 2^{2^3} + 1 \equiv 4^2 + 2^2 + 1 &\equiv 2 + 4 + 1 \pmod7,\\
\end{align}
and so forth.
You should now know the congruence classes of $4^{2^n}$ and $2^{2^n}$ for every $n$ and be able to prove the obvious pattern of congruences for alternating odd and even $n$ by induction.
This is just a little less elegant that some of the other proofs in that you may find it necessary to have two cases in the inductive step, one for odd $k$ and one for even $k.$
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