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

[Math100]

Homework Statement

For $n\geq 1$, use congruence theory to establish the following divisibility statement:
$13\mid 3^{n+2}+4^{2n+1}$.

Relevant Equations

None.

Proof

Let $n\geq 1$ be a natural number.
Then \begin{align*} 3^{n+2}+4^{2n+1}&\equiv 3^{n}\cdot 3^{2}+(4^{2})^{n}\cdot 4\pmod {13}\\
&\equiv (3^{n}\cdot 9+16^{n}\cdot 4)\pmod {13}\\
&\equiv (3^{n}\cdot 9+3^{n}\cdot 4)\pmod {13}\\
&\equiv (3^{n}\cdot 13)\pmod {13}\\
&\equiv 0\pmod {13}.
\end{align*}
Therefore, $13\mid 3^{n+2}+4^{2n+1}$ for $n\geq 1$.

 

[Mark44]

Homework Statement:: For $n\geq 1$, use congruence theory to establish the following divisibility statement:
$13\mid 3^{n+2}+4^{2n+1}$.
Relevant Equations:: None.

Proof

Let $n\geq 1$ be a natural number.
Then \begin{align*} 3^{n+2}+4^{2n+1}&\equiv 3^{n}\cdot 3^{2}+(4^{2})^{n}\cdot 4\pmod {13}\\
&\equiv (3^{n}\cdot 9+16^{n}\cdot 4)\pmod {13}\\
&\equiv (3^{n}\cdot 9+3^{n}\cdot 4)\pmod {13}\\
&\equiv (3^{n}\cdot 13)\pmod {13}\\
&\equiv 0\pmod {13}.
\end{align*}
Therefore, $13\mid 3^{n+2}+4^{2n+1}$ for $n\geq 1$.

Looks good.

 

[fresh_42]

Homework Statement:: For $n\geq 1$, use congruence theory to establish the following divisibility statement:
$13\mid 3^{n+2}+4^{2n+1}$.
Relevant Equations:: None.

Proof

Let $n\geq 1$ be a natural number.
Then \begin{align*} 3^{n+2}+4^{2n+1}&\equiv 3^{n}\cdot 3^{2}+(4^{2})^{n}\cdot 4\pmod {13}\\
&\equiv (3^{n}\cdot 9+16^{n}\cdot 4)\pmod {13}\\
&\equiv (3^{n}\cdot 9+3^{n}\cdot 4)\pmod {13}\\
&\equiv (3^{n}\cdot 13)\pmod {13}\\
&\equiv 0\pmod {13}.
\end{align*}
Therefore, $13\mid 3^{n+2}+4^{2n+1}$ for $n\geq 1$.

Perfect. Only a minor thing: It is generally better to group things that belong together with parentheses. So $13\mid (3^{n+2}+4^{2n+1})$ would be better in my opinion. It is not important in this case since everybody can see that $13\nmid 3^m$ but it is good to get used to it for any linear notation.