Prove 43 divides 6^(n+2)+7^(2n+1) for all n≥1

[Math100]

Homework Statement

For $n\geq 1$, use congruence theory to establish the following divisibility statement:
$43\mid (6^{n+2}+7^{2n+1})$.

Relevant Equations

None.

Proof:

Let $n\geq 1$ be a natural number.
Then \begin{align*} 6^{n+2}+7^{2n+1}&\equiv (6^{n}\cdot 6^{2}+(7^{2})^{n}\cdot 7)\pmod {43}\\
&\equiv (6^{n}\cdot 36+49^{n}\cdot 7)\pmod {43}\\
&\equiv (6^{n}\cdot 36+6^{n}\cdot 7)\pmod {43}\\
&\equiv (6^{n}\cdot 43)\pmod {43}\\
&\equiv 0\pmod {43}.
\end{align*}
Therefore, $43\mid (6^{n+2}+7^{2n+1})$ for $n\geq 1$.

 

[fresh_42]

Yep. And another remark concerning the parentheses.

The "divides" thingy and the addition actually are distributive, and if I remember correctly, you already (correctly) used it in another thread. I mean the other direction: $n\,|\,a\wedge n\,|\,b\Longrightarrow n\,|\,(a+b).$