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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?

Bill Dubuque
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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
  • That's a dupe of this because exactly the same type of covering xongruence argument works there - as is explained in the closing comment there. Once you know this idea it is rote arithmetic to apply it. – Bill Dubuque Jun 22 '24 at 22:22

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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The first $\,\color{#c00}4\,$ elements of $\,f_K\! = 19\cdot 8^{K}\!+17\,$ have these prime factorizations

$\!\begin{align} f_0 = 19\cdot 8^0+17 &= 2^2\, \color{#0af}3^2\\ f_1= 19\cdot 8^1+17 &= \color{#90f}{13}^2\\ f_2 =19\cdot 8^2+17 &= \color{#0af}3^2\, 137\\ f_3 = 19\cdot 8^3+17 &= \color{darkorange}5\ 1949 \end{align}$

and $\,8^{\large\color{#c00}4}\!-1 = \color{#0af}3^2\cdot \color{darkorange}5\cdot 7\cdot \color{#90f}{13} = \prod p_i ,\,$ where we see for all $\,K <\color{#c00} 4,\,$ $\rm\color{#0af}{that}\ \color{darkorange}{some}\ \color{#90f}{prime}$ $\,p_i\mid f_K\,$ which implies that for all $\,N\ge 0\,$ some $\,p=p_i\mid f_N,\,$ so $\,f_N\,$ is composite for all $\,N\ge 0,\,$ by

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

$\!\!\!\bmod p\!:\ \color{#c00}{8^{4}}\equiv\color{#c00} 1\Rightarrow \color{#0a0}{8^N}\! \equiv 8^{N\bmod\color{#c00}4}\!\equiv \color{#0a0}{8^K},\, K<\color{#c00}4,\,$ 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.

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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