exponential function - Use theory of congruence to establish divisibility

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 $43$

Answer

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 $43.$