<s>[INST] Show using induction that for every $n\in \mathbb{N}$ the number $P_n=2^{2^{n+1}}+2^{2^n}+1$ is divisible by $21$. Show using induction that for every $n\in \mathbb{N}$ the number $P_n=2^{2^{n+1}}+2^{2^n}+1$ is divisible by $21$.
So the first step is to check for $n=1$ $$P_1=2^{2^2}+2^2+1=16+5=21,$$ which is divisible by $21$, so the statement holds for $n=1$. Let for some $n=k\ge1,P_k=2^{2^{k+1}}+2^{2^k}+1$ is divisible by $21$. Now we should try to prove that the statement also holds for $n=k+1$, or show that $P_{k+1}=2^{2^{k+2}}+2^{2^{k+1}}+1$ is divisible by $21$. We can write $P_{k+1}$ as $$P_{k+1}=2^{2\cdot2^{k+1}}+2^{2\cdot2^k}+1$$ How do we use the induction hypothesis?
 [/INST] Let $A_n=2^{2^n}$. Then $P_n=2^{2^{n+1}}+2^{2^n}+1=A_n^2+A_n+1$ and...
$$P_{n+1}=2^{2^{n+2}}+2^{2^{n+1}}+1=2^{4\cdot2^n}+2^{2\cdot2^n}+1=A_n^4+A_n^2+1\\=(A_n^2+A_n+1)(A_n^2-A_n+1)=P_n(A_n^2-A_n+1)$$
And $P_1=21$. So... the answer is clear! If $21\mid P_n$, then automatically $21\mid P_{n+1}$ holds!
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