19⋅8n+17$19\cdot 8^n+17$ is composite for all natural n$n$ (covering congruences)

If n≡1(mod4)$n\equiv 1(\mathrm{mod}4)$, then 8n≡8(mod13)$8^n\equiv 8(\mathrm{mod}13)$ and so

If n≡3(mod4)$n\equiv 3(\mathrm{mod}4)$, then 8n≡2(mod5)$8^n\equiv 2(\mathrm{mod}5)$ and so

If n$n$ is even, then 19⋅8n+17$19\cdot 8^n+17$ is divisible by 3$3$.

And clearly 19⋅8n+17>13,∀n∈Z+$19\cdot 8^n+17>13,\mathrm{\forall }n\in {\mathbb{Z}}^{+}$.

The first 4$4$ elements of fK=19⋅8K+17$f_K=19\cdot 8^K+17$ have these prime factorizations

f0=19⋅80+17f1=19⋅81+17f2=19⋅82+17f3=19⋅83+17=2232=132=32137=5 1949$\begin{array}{rl}{f}_0=19\cdot 8^0+17& =2^2{3}^2\\ f_1=19\cdot 8^1+17& ={13}^2\\ f_2=19\cdot 8^2+17& =3^{2}137\\ f_3=19\cdot 8^3+17& =51949\end{array}$

and 84−1=32⋅5⋅7⋅13=∏pi,$8^4-1=3^2\cdot 5\cdot 7\cdot 13=\prod p_i,$ where we see for all K<4,$K<4,$ that some prime$\mathrm{t}\mathrm{h}\mathrm{a}\mathrm{t}\mathrm{s}\mathrm{o}\mathrm{m}\mathrm{e}\mathrm{p}\mathrm{r}\mathrm{i}\mathrm{m}\mathrm{e}$ pi∣fK$p_i\mid f_K$ which implies that for all N≥0$N\ge 0$ some p=pi∣fN,$p=p_i\mid f_N,$ so fN$f_N$ is composite for all N≥0,$N\ge 0,$ by

p∣fK⇒0≡19⋅8K+17≡19⋅8N+17≡fN,$p\mid f_K\Rightarrow 0\equiv 19\cdot 8^K+17\equiv 19\cdot 8^N+17\equiv f_N,$ so p∣fN,$p\mid f_N,$ properly by p<f4≤fN,$p<f_4\le f_N,$ by

modp: 84≡1⇒8N≡8Nmod4≡8K,K<4,$modp:8^4\equiv 1\Rightarrow 8^N\equiv 8^{Nmod4}\equiv 8^K,K<4,$ by mod order reduction.

Remark $$ This is a prototypical application of covering congruences. See also this question, and see this question for a polynomial analog, and a link to a paper of Schinzel.

Not sure whether this qualifies for it as answer but if you reduce it mod 3$3$ you can see that the expression is divisible by 3$3$ if n$n$ is even.

If you reduce it mod 13$13$ then you see that the expression is divisible by 13$13$ if n≡1$n\equiv 1$ mod 4$4$. (84≡1$8^4\equiv 1$ mod 13$13$ )

If you reduce it mod 5$5$ then the expression is divisible by 5$5$ if n$n$ is 3$3$ mod 4$4$.

I sort of calculated a few numbers and then tried some numbers. I don't think this is a good way at all but I guess it works.