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On a Google search, I've notice that often the "trick" involves factoring the expression (and thus showing that the expression isn't prime), but I can't see it. Is this how you would go about it in this case, or would that be a dead end? Any hints?

Gordon
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  • How is $19 * 8^n + 17 $ a duplicate of $2^n + 1518781$ ???

    https://math.stackexchange.com/questions/4841832/2n-1518781-is-never-a-prime-why

     – mick Jan 11 '24 at 11:47

3 Answers3

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If $n\equiv 1\pmod {4}$, then $8^n\equiv 8\pmod {13}$ and so $$19\cdot 8^n+17\equiv 19\cdot 8 + 17\equiv 0\pmod {13}$$

If $n\equiv 3\pmod{4}$, then $8^n\equiv 2\pmod {5}$ and so $$19\cdot 8^n + 17\equiv 19\cdot 2 + 17\equiv 0\pmod {5}$$

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

And clearly $19\cdot 8^n + 17>13, \forall n\in\mathbb Z^+$.

user26486
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5

Hint $ $ Looking at the prime factorizations of the first $\color{#c00}4$ elements we notice the following.

$ 8^{\large\color{#c00}4}\!-1 = 3^2\cdot 5\cdot 7\cdot 13 = \prod p_i ,\,$ and some prime $\,p_i\mid f_K\! = 19\cdot 8^{K}\!+17\,$ for all $\,K <\color{#c00} 4,\,$ therefore some $\,p_i\mid f_N\,$ for all $\,N,\,$ so $\,f_N\,$ is composite for all $\,N\ge 0.\,$

Proof $\ $ Dividing $\,N\,$ by $\,4\,$ yields $\,N = K\!+4J\,$ for $\,K < 4.\ $ By hypothesis some $\,p = p_i\mid f_K$

${\rm mod}\ p\!:\,\ \color{#c00}{8^{ 4}}\equiv\color{#c00} 1\,\Rightarrow\, \color{#0a0}{8^N} \equiv 8^{K+4J}\!\equiv 8^K(\color{#c00}{8^4})^J\equiv 8^K\color{#c00}1^J\equiv \color{#0a0}{8^K}\ $ by $ $ Congruence Rules.

$p\mid f_K\Rightarrow\, 0\equiv 19\cdot \color{#0a0}{8^K}\!+17\equiv 19\cdot \color{#0a0}{8^N}\!+17 \equiv f_N\Rightarrow p\mid f_N,\,$ properly by $\,p < f_4 \le f_N$

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.

Bill Dubuque
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  • See here for another example. – Bill Dubuque Feb 17 '15 at 21:06
  • Doubts: (a) How do we figure out the number of elements $\color{red}{n}$ to factorize before noticing a pattern, and (b) why do we look at $8^{\color{red}{n}} - 1$ when searching for the factors of $f_N$. – Anant Feb 18 '15 at 18:38
  • @Anant Keep increasing $,n,$ till the first $,n,$ elements all have a factor from $,8^n-1.,$ If the compositeness can be proved via such a congruential case-analysis then this method will eventually discover it. Of course, without any a priori bound, it is only a semi-algorithm. But it works well for "designed" problems, where we expect this may occur. – Bill Dubuque Feb 18 '15 at 18:50
  • Very nice! It will take me some time to understand the full content of the paper, but I'm already liking the idea of covering congruences :) – Anant Mar 02 '15 at 18:27
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Not sure whether this qualifies for it as answer but if you reduce it mod $3$ you can see that the expression is divisible by $3$ if $n$ is even.

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

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

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.

Jack Yoon
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