TITLE: Proving that $19\mid 5^{2n+1}+3^{n+2} \cdot 2^{n-1}$
QUESTION [5 upvotes]: 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?

REPLY [5 votes]: Its arithmetical essence is clarified using Congruence Product and Power Rules, namely
$$\begin{align}{\rm mod}\,\ 19\!:\qquad \big[\,\  5^{\large\color{#c00}1} \ &\equiv\, -3^{\large\color{#c00} 2}\ \cdot\, 2^{\large\color{#c00}{-1}}\big]\ \ \ {\rm by}\ \ \ \ \,2\cdot 5\equiv -3^{\large 2}\\
 \times\ \ \, \big[\, 5^{\large 2n} &\equiv\ \ \  3^{\large n}\,\cdot\, 2^{\large n}\ \ \ \big]\ \ \ {\rm by} \ \ \Big[\,5^{\large 2}\equiv 3\cdot 2\,\Big]^{\large n}\\[.1em]
\Rightarrow\,\ 5^{\large 2n+\color{#c00}1}\!&\equiv -3^{\large n+\color{#c00}2} 2^{\large n\color{#c00}{-1}}
\end{align}\qquad\qquad\quad\ $$
Therefore we infer that $\ 19\mid 5^{\large 2n+1}+\ 3^{\large n+2} 2^{\large n-1}$ when it is integral, i.e. for all $\,n\ge 1$
Remark $ $ It is even clearer when written in fractional form, namely
$$ \dfrac{5^{\large 2n+1}}{3^{\large n+2}2^{\large n-1} }\ \equiv\ \left[\dfrac{2\cdot 5}{3\cdot 3}\right]\left[\dfrac{5^{\large 2}}{3\cdot 2}\right]^{\large n}\!\!\equiv\,  -1\cdot 1^{\large n}\equiv\, -1$$
Notice how use of congruence language greatly simplifies the inductive step, reducing it to the trivial induction that $\,1^n\equiv 1.$