I had this problem. I solved it. tell me if it is correct I have to prove $6^{n+1} +7^{2n+1}$ is divisible by $43$ when $n\geq 1$ My solution $6^{n+1} +7^{2n+1}$=$216.6^{n-1}+343.49^{n-1}$
$\equiv 1.6^{n-1} +(-1).6^{n-1}$ $since 343 \equiv -1, 216\equiv 1, 49\equiv 6 (mod 43)$
$\equiv 6^{n-1}-6^{n-1}$ $\equiv 0$ since $n\geq 1, 6^{n-1}-6^{n-1}$ is an integer
Since $6^{n+1} +7^{2n+1} \equiv 0 (mod 43)$ so it is divisible by $43$
I think your statement is wrong.
For $n=1$ we obtain $379$ is divisible by $43$, which is wrong.
By the way, for all natural $n$ we obtain: $$6^{n+1}+7^{2n-1}=36\cdot6^{n-1}+7\cdot49^{n-1}=7\left(49^{n-1}-6^{n-1}\right)+43\cdot6^{n-1}$$ is divisible by $43.$