Exercise 2.1.14 ($7 \mid (3^{2n+1} + 2^{n+2})$ for all $n$)

Question

Show that $7 \mid (3^{2n+1} + 2^{n+2})$ for all $n$.

Richard Ganaye · Accepted Answer

\begin{proof}
 Since $9 \equiv 2 \pmod 7$,
\begin{align*}
3^{2n+1} + 2^{n+2} &=  3\cdot 9^n + 4\cdot 2^n\
&\equiv 3 \cdot 2^n + 4 \cdot 2^n = 7 \cdot 2^n \equiv 0 \pmod 7.
\end{align*}
Therefore 
$$7 \mid 3^{2n+1} + 2^{n+2}$$ for all integer $n$.

\end{proof}

Exercise 2.1.14 ($7 \mid (3^{2n+1} + 2^{n+2})$ for all $n$)

Answers

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Exercise 2.1.14 ($7 \mid (3^{2n+1} + 2^{n+2})$ for all $n$)

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