Prove that: $3^{3n - 2} + 2^{3n + 1}$ is a multiple of 19

Question

Let $~n~$ be any natural number. Prove that: $$3^{3n - 2} + 2^{3n + 1}$$ is a multiple of $~19~$ .

Answer 1

Hint: One way is to proceed by induction, the base case is trivial. Assume for $n=k$ we have that $3^{3k-2}+2^{3k+1}=19p$ for some $p\in \mathbb{Z}$. Then, we have that for $n=k+1$: $$3^{3(k+1)-2}+2^{3(k+1)+1}=3^3\cdot 3^{3k-2}+2^3\cdot 2^{3k+1}=\cdots$$

Answer 2

Hint: It is $$ 3^{3n - 2} + 2^{3n + 1}= \frac{27^n}{9}+8^n\cdot 2\equiv \frac{8^n}{9}+8^n\cdot 2 \mod 19= 8^n\cdot \frac{19}{8} \mod 19 = 8^{n-1}\cdot 19 \mod 19 $$

Answer 3

As $3^3\equiv2^3\pmod{19}$

$$3\cdot(3^3)^{n-1}+2^{3n+1}\equiv3\cdot(2^3)^{n-1}+2^{3n+1}$$

$$\equiv2^{3n-3}(3+2^4)\equiv0\pmod{19}$$

Answer 4

For $\,n=k\!+\!1\,$ it is $\,3\, \overbrace{\color{#c00}3^{\large \color{#c00}3k}}^{\Large\color{#c00} 2^{\LARGE\color{#c00} 3k}}\!+2^{\large 4}\ 2^{\large 3k} \equiv 19\, 2^{\large 3k}\equiv 0\ $ by $\,\color{#c00}{3^{\large 3}\equiv 2^{\large 3}}\pmod{\!19}$

Alternatively its $\equiv 0\!\iff\! 3^{\large 3n}/9 \equiv - 2\,2^{\large 3n }\!\iff\! (\color{#c00}{3/2})^{\large\color{#c00} 3n}\equiv -18\equiv 1,\,$ true by $\rm\color{#c00}{above}$.