- #1

Math100

- 780

- 220

- 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 ##.

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 ##.