<s>[INST] Prove that $3\cdot 5^{2n+1} +2^{3n+1}$ is divisible by $17$ for all $n ∈ \mathbb{N}$ 
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.
 [/INST] Here are two alternative routes.
Using the binomial theorem:
$$
3\cdot 5^{2n+1} +2^{3n+1}
= 15\cdot 25^{n} +2\cdot 8^{n}
= 15\cdot (17+8)^{n} +2\cdot 8^{n}\\
= 15\cdot (17a+8^{n}) +2\cdot 8^{n}
= 17b+15\cdot 8^{n} +2\cdot 8^{n}
= 17b+17\cdot 8^{n}
$$
Using recurrences:
Let
$
x_n=3\cdot 5^{2n+1} +2^{3n+1}
$,
Since, as above,
$
x_n= 15\cdot 25^{n} +2\cdot 8^{n}
$,
we have the recurrence $x_{n}=33x_{n-1}-200x_{n-2}$, for $n\ge 2$ (*).
The result follows by induction because $x_0$ and $x_1$ are multiples of $17$.
(*) In general, if $x_n=a\alpha^{n} +b\beta^{n}$, then $x_{n}=(\alpha+\beta)x_{n-1}-(\alpha\beta)x_{n-2}$, for $n\ge 2$. This follows because $\alpha$ and $\beta$ are the roots of the quadratic $(x-\alpha)(x-\beta)=x^2-(\alpha+\beta)x+(\alpha\beta)$.
</s>