TITLE: Prove that $3\cdot 5^{2n+1} +2^{3n+1}$ is divisible by $17$ for all $n ∈ \mathbb{N}$
QUESTION [5 upvotes]: Use mathematical induction to prove that $3\cdot 5^{2n+1} +2^{3n+1}$
  is divisible by $17$ for all $n ∈ \mathbb{N}$.

I've tried to do it as follow. 
If $n = 1$ then  $392/17 = 23$.
Assume it is true when $n = p$. Therefore $3\cdot 5^{2p+1} +2^{3p+1} = 17k $ where $k ∈ \mathbb{N} $. Consider now $n=p+1$. Then 
\begin{align}
&3\cdot 5^{2(p+1)+1} +2^{3(p+1)+1}=\\
&3\cdot 5^{2p+1+2} + 2^{3p+1+3}=\\
&3\cdot5^{2p+1}\cdot  5^{2} + 2^{3p+1}\cdot 2^{3}.
\end{align}
I reached a dead end from here. If someone could help me in the direction of the next step it would be really helpful. Thanks in advance.

REPLY [4 votes]: Conceptually the inductive step scales by $\,\color{#c00}{5^{\large 2} \equiv\, 2^{\large 3}}\pmod{\!17}.\,$ If you don't know congruences we can preserve this arithmetical essence by using a divisibility (vs. congruence) product rule where we used standard notation $\ m\mid n\ $ means $\,m\,$ divides $\,n.\,$
$$\qquad\ \  \begin {align}
&17\mid\ \ \ \color{#c00}{5^{\large 2}\quad\ \ -\quad\ \  2^{\large 3}}\\
&17\mid\ a\ \ \ 5^{\large 2n+1} -\ b\ \ \ \,2^{\large 3n+1}\qquad\ P(n)\\
\Rightarrow\ \ &17\mid a \color{#c00}{5^{\large 2}} 5^{\large 2n+1}\! -\,\ b \color{#c00}{2^{\large3}} 2^{\large 3n+1}\qquad P(n\!+\!1)
\end{align} $$
$\begin{align}{\bf Divisibility\ Product\ Rule}\ \ \ \  
&m\mid\  a\ -\ b\qquad {\rm i.e.}\quad \   a\,\equiv\, b\\
&m\mid \ \ A\: -\: B\qquad\qquad \ A\,\equiv\, B\\
\Rightarrow\ \ &\color{}{m\mid aA - bB}\quad  \Rightarrow\quad aA\equiv bB\!\pmod{\!m}\\[.2em]
{\bf Proof}\,\ \  m\mid (\color{#0a0}{a\!-\!b})A + b(\color{#0a0}{A\!-\!B}) &\,=\, aA-bB\ \ \text{by $\,m\,$ divides $\rm\color{#0a0}{green}$ terms by hypothesis.}\end{align}$
You can find further discussion in many prior posts.