- #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 ##.