number theory - Proving that $19mid 5^{2n+1}+3^{n+2} cdot 2^{n-1}$

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?

Answer

You can prove this by induction.

---

First, show that this is true for $n=1$:

$5^{2\cdot1+1}+3^{1+2}\cdot2^{1-1}=19\cdot8$

Second, assume that this is true for $n$:

$5^{2n+1}+3^{n+2}\cdot2^{n-1}=19k$

Third, prove that this is true for $n+1$:

$5^{2(n+1)+1}+3^{n+1+2}\cdot2^{n+1-1}=$

$5^{2+2n+1}+3^{1+n+2}\cdot2^{1+n-1}=$

$5^{2}\cdot5^{2n+1}+3^{1}\cdot3^{n+2}\cdot2^{1}\cdot2^{n-1}=$

$5^{2}\cdot5^{2n+1}+3^{1}\cdot2^{1}\cdot3^{n+2}\cdot2^{n-1}=$

$25\cdot5^{2n+1}+\color\green{6}\cdot3^{n+2}\cdot2^{n-1}=$

$25\cdot5^{2n+1}+(\color\green{25-19})\cdot3^{n+2}\cdot2^{n-1}=$

$25\cdot5^{2n+1}+25\cdot3^{n+2}\cdot2^{n-1}-19\cdot3^{n+2}\cdot2^{n-1}=$

$25\cdot(\color\red{5^{2n+1}+3^{n+2}\cdot2^{n-1}})-19\cdot3^{n+2}\cdot2^{n-1}=$

$25\cdot\color\red{19k}-19\cdot3^{n+2}\cdot2^{n-1}=$

$19\cdot25k-19\cdot3^{n+2}\cdot2^{n-1}=$

$19\cdot(25k-3^{n+2}\cdot2^{n-1})$

---

Please note that the assumption is used only in the part marked red.