<s>[INST] How to prove by induction that $4^{2n}-3^{2n}-7$ is divisible by 84 for any n starting from 1? How to prove by induction that $4^{2n}-3^{2n}-7$ is divisible by 84 for any n starting from 1 ?
Take n=1 and prove for the base case, assume its true for some n, then the third step went like this: $$4^{2n+2}-3^{2n+2}-7=84k=4^24^{2n}-3^23^{2n}-7=84k=4^2(4^{2n}-3^{2n}-7)+7*3^{2n}+105$$ After using the assumption the expression becomes $$4^284k+63*3^n+105=84k$$ and I'm stuck here, I can't set $63*3^n+105$ to factor with 84.
 [/INST] for $n=1$ we have $4^{2}-3^{2}-7=0$ and $0$ is divisible by $84$.
Let $4^{2k}-3^{2k}-7$ is divisible by $84$. We show $4^{2k+2}-3^{2k+2}-7$ is divisible by $84$.
$$4^{2k}-3^{2k}-7=84q$$
$$4^{2k+2}-3^{2k}(9+7)-7\times 16=84(16q)$$
$$4^{2k+2}-3^{2k+2}-7=84(16q+1)+7(3+3^{2k})$$
$$4^{2k+2}-3^{2k+2}-7=84(16q+1)+21(1+3^{2k-1})$$
$$4^{2k+2}-3^{2k+2}-7=84(16q+1)+21(4q')$$
$$4^{2k+2}-3^{2k+2}-7=84(16q+1)+84q'$$
Note
$${{3}^{2k-1}}+1\overset{4}{\mathop{\equiv }}\,0$$
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