<s>[INST] Proving that $19\mid 5^{2n+1}+3^{n+2} \cdot 2^{n-1}$ How can I prove that $$5^{2n+1}+3^{n+2} \cdot 2^{n-1} $$ can be divided by 19 for any nonnegative n? What modulo should I choose?
 [/INST] Denote $\mathcal{P}(n)$ the statement that $5^{2n+1} + 3^{n+2}\cdot 2^{n-1}$ is divisible by $19$. You can check for yourself that $\mathcal{P}(1)$ is true. A setup for the proof of $\mathcal{P}(n+1)$:
\begin{align} 
5^{2(n+1)+1} + 3^{(n+1)+2}\cdot 2^{(n+1)-1} & = \\
25 \times 5^{2n+1} + 6 \times 3^{n+2}\cdot 2^{n-1}  & = \\
19 \times 5^{2n+1} + 6 \times 5^{2n+1} + 6 \times 3^{n+2}\cdot 2^{n-1} & = \cdots
\end{align} can you finish the proof from here?
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