Prove that $5$ divides $3^{3n+1}+2^{n+1}$ 
Prove that $5$ divides $3^{3n+1}+2^{n+1}$

I tried to prove the result by induction but I couldn't.
The result is true for $n=1$.
Suppose that the result is true for $n$ i.e $3^{3n+1}+2^{n+1}=5k$ for some $k\in \mathbb{N}$. We study the term 
$$3^{3n+4}+2^{n+2}=3^{3n+1}3^3+2^{n+1}2$$
I tried to prove that that the difference is a multiple of $5$.
$$3^{3n+1}3^3+2^{n+1}2-3^{3n+1}+2^{n+1}=2(3^{3n+1}\cdot 13+2^n)$$
Therefore it's enough to prove that $3^{3n+1}\cdot 13+2^n$ is a multiple of $5$. But if I do again this method applied to this "new problem" is get something similar. I think that there exist a different method to do this using induction.
 A: Your approach works if you just subtract once more:
$$
3^{3n+4} + 2^{n+2} - 2\cdot 5k\\
= 27\cdot 3^{3n+1} + 2\cdot 2^{n+1} - 2(3^{3n+1} + 2^{n+1})\\
= 25\cdot 3^{3n+1}
$$
and you're done.
A: $3 (27^{n})+2 (2^{n}) $ is equal to your value,so $3 (2^{n})+2 (2^{n})=5( 2^{n}) $ has equal reminder (as dividing by 5)with the previous value because 27 has the same reminder ( with respect to 5) as 2.so it is proved.
A: I don't see why people want to avoid modular arithmetic in solving problems like this. Fermat's little theorem is not required:
Let $f(n) = 3^{3n+1}$ and $g(n) = 2^{n+1}$. We want to prove that $f(n) \equiv -g(n) \;(\mathrm{mod}\;5)$. When $n=0$ this says that $3 \equiv -2\;(\mathrm{mod}\;5)$ which is true. Noting that $f(n+1) = 27f(n) \equiv 2f(n) \;(\mathrm{mod}\;5)$ and $g(n+1) = 2g(n)$, we have that, if $f(n) \equiv -g(n) \;(\mathrm{mod}\;5)$, then also
$$f(n+1) \equiv 2f(n) \equiv -2g(n) \equiv -g(n+1) \;(\mathrm{mod}\;5)$$
and so what we wanted to prove follows by induction.
A: Hint :
$$3^{3n+4}+2^{n+2}=27\times 3^{3n+1}+2\times 2^{n+1}=2\times(3^{3n+1}+2^{n+1})+25\times 3^{3n+1}$$
A: $$3^{3n+4}+2^{n+2}=3^{3n+1}\cdot 3^{3} +2^{n+1}\cdot 2\\
=(5k-2^{n+1})\cdot 3^{3}+2^{n+1}\cdot 2\\
=2^{n+1}(2-27)+5k\cdot 27=-2^{n+1}\cdot 25+5k\cdot 27\\
=5(-2^{n+1}\cdot 5+27k)$$ which is divisible by $5$.
A: Or we could just expand and rearrange$$
\begin{align}
3^{3n+1}+2^{n+1} 
&= 3\cdot27^n + 2\cdot2^n \\
&= 3\cdot(25+2)^n +  2\cdot2^n  \\
&= 3\left(5k+2^n \right) + 2\cdot 2^n  \tag{Using Binomial Theorem}\\
&= 5\cdot k'+3\cdot2^n+2\cdot2^n  \\
&= 5\cdot k'+5\cdot 2^n \\
\end{align}
$$
Hence proved
A: try this
$$3^{3n+1}3^3+2^{n+1}2-3^{3n+1}+2^{n+1}=3^{3n+1}\cdot26+2^{n+1}$$
$$3^{3n+1}\cdot26+2^{n+1}=3^{3n+1}\cdot25+3^{3n+1}+2^{n+1}$$
since $3^{3n+1}+2^{n+1}=5k$
$$3^{3n+1}\cdot25+3^{3n+1}+2^{n+1}=3^{3n+1}\cdot25+5k$$
the difference is divisable by $5$
