<s>[INST] Prove by induction that $3^{3n+1} + 2^{n+1}$ is divisible by 5 How do I do this? I've tried using logarithms, factoring, but nothing seems to work.
 [/INST] HINT:
If $f(n)=3^{3n+1}+2^{n+1}$
$$f(m+1)-a\cdot f(m)=3^{3m+1}(3^3-a)+2^{m+1}(2-a)$$
If we set 
$3^3-a=0\iff a=3^3\implies f(m+1)-3^3f(m)=\cdots\equiv0\pmod5$
or
$2-a=0\iff a=2\implies f(m+1)-2f(m)=\cdots\equiv0\pmod5$
$\implies5|f(m+1)\iff5|f(m)$

Non-inductive methods:
$(1)$Binomial Expansion:
$3^3=25+2\implies3^{3n+1}+2^{n+1}=2^{n+1}+3(2+25)^n$
$=2^{n+1}+3[2^n+\sum_{r=1}^n\binom nr2^{n-1}25^r]=2^n(2+3)+($ a multiple of $25)$
$(2)$Modular Arithmetic 
$3^3\equiv2\pmod5\implies3^{3n+1}=3(3^3)^n\equiv3\cdot2^n\pmod5$
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