<s>[INST] Is it a composite number? How do I prove $19\cdot8^n+17$ is a composite number? 
Or is that number just a prime?
So I tried to find a divisor in the cases $ n = 2k $ and $ n = 2k + 1 $. But I had no success.
Do you have any ideas?
 [/INST] Three cases:  


*

*$n$ is even. Then $$19\cdot 8^n+17\equiv 1\cdot (-1)^n+17\equiv 1+17\equiv  0\pmod{\! 3}$$

*$n=4k+1$ for some $k\in\Bbb Z_{\ge 0}$. Then $$19\cdot 8^{4k+1}+17\equiv 6\cdot \left(8^2\right)^{2k}\cdot
   8+4\equiv 6\cdot (-1)^{2k}\cdot 8+4$$
$$\equiv 48+4\equiv 52\equiv 0\pmod{\! 13}$$ 


*

*$n=4k+3$ for some $k\in\Bbb Z_{\ge 0}$. Then $$19\cdot 8^{4k+3}+17\equiv (-1)\cdot \left(8^2\right)^{2k}\cdot 8^3+17\equiv (-1)\cdot (-1)^{2k}\cdot 8^3+17$$


$$\equiv -(-2)^3+17\equiv 8+17\equiv 25\equiv 0\pmod{\! 5}$$
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