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?

$5^{2n+1} = 5 \times 25^n \equiv 5 \times 6^n$ (modulo 19).
Hence we have $5^{2n+1} + 3^{n+2}\times2^{n-1} \equiv 6^{n-1} \times (30 + 27) = 6^{n-1}\times3\times 19$ (modulo 19)
