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.
On
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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From your work, with $f(n)=4^{2n}-3^{2n}-7:\;$ If $\;84|f(n) , $ then $$(\;84|f(n+1)\iff 84|(7\cdot 3^{2n}+105)\;).$$ $$\text {For } n\geq 1 \text { we have }\quad 7\cdot 3^{2n}+105=21(3^{2n-1}+5).$$ You can confirm by induction (or other means) that $ 4|(3^{2n-1}+5) , $ so $ 84 | (21(3^{2n-1}+5)).$
You have $7\cdot 3^{2n} + 105 = 7(3^{2n} + 15)$ which you want to show is divisible by $84 = 7\cdot 12$. That amounts to showing that $3^{2n} + 15$ is divisible by $12$. It is obviously divisible by $3$, so it remains to show that it is divisible by $4$. Now, $3 = 4-1$, so we may use the binomial theorem: $$ 3^{2n} = (4-1)^{2n} = \sum_{i = 0}^{2n}\binom{2n}{i}4^i(-1)^{2n-i} $$ Note that the only term in that sum that's not divisible by $4$ is a $1$, so $3^{2n} = 4l + 1$ for some $l$. Thus $$ 3^{2n} + 15 = 4l + 1 + 15 = 4l + 16 $$ is divisible by $4$, and we're done.