For ## n\geq 1 ##, use congruence theory to establish....

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For any natural number n ≥ 1, it is established that 43 divides the expression 6^(n+2) + 7^(2n+1) using congruence theory. The proof shows that the expression simplifies to 0 modulo 43, confirming the divisibility. Additionally, the discussion highlights the distributive property of divisibility, noting that if n divides both a and b, then n also divides their sum. This reinforces the mathematical principles applied in the proof. The conclusion is that the relationship holds true for all natural numbers n starting from 1.
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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 ##.
 
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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).##
 
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