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Another question that I can only solve in part.

Is there an $n$ such that $324+455^n$ is prime?

When $n$ is odd, this is false since

$$ 324+455^n = (2\cdot 3^2)^2+(5\cdot 91)^n \equiv (-1)^2+(5\cdot 15)^n \equiv 1+(6\cdot 3)^n \equiv 1+(-1)^n\pmod{19}$$

so that $19\mid 324+455^n$ whenever $n$ is odd. Now, what about $n$ even?

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    $\begingroup$ Kind of related math.stackexchange.com/questions/597234/… $\endgroup$
    – snulty
    Apr 19 '14 at 21:09
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    $\begingroup$ While one quckly checks that $n\equiv 2\pmod 4$ produces a multiple of $17$, such modular argumentation seems to become moot for the remaining cases. With $n=28$, the smallest prime factor is $13553$, and $f(4),f(8),f(12),f(16),f(20),f(24),f(28)$ are pairwise(!) coprime. $\endgroup$ Apr 19 '14 at 21:09
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    $\begingroup$ assuming I didn't mess up with my J code there are no primes for $n \le 500$ $\endgroup$ Apr 19 '14 at 21:10
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Let's analyse three possible cases:

If $n\equiv 1\pmod 2$, we have $455^n+324 \equiv 0\pmod {19}$ because

$$\begin{eqnarray} (455^1+324) & = & 19\times 41 \\ (455^2-1) & = & 19\times 10896\\ \end{eqnarray}$$

If $n\equiv 2\pmod 4$, we have $455^n+324 \equiv 0\pmod {17}$ because $$\begin{eqnarray} (455^2+324) & = & 17\times 12197\\ (455^4-1) & = & 17\times 2521138272\\ \end{eqnarray}$$

Finally, if $n=4k$, we can apply the equality $$x^4+4y^4 = (x^2 + 2xy + 2y^2)(x^2 - 2xy + 2y^2)$$ to $x=455^k$ and $y=3$ to obtain $$455^{4k}+4\cdot3^4 = (455^{2k} + 6\cdot 455^k + 18)(455^{2k}-6\cdot 455^k + 18)$$

Since both factors are strictly greater than $1$, their product is certainly a composite number.

Thus, the quantity $455^n+324$ is not a prime for any natural number $n$.

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  • $\begingroup$ Though I don't see yet how you go from $(455^1+324)$ and $(455^2-1)$ to arbitrary $(455^n+324)$, for odd $n$. $\endgroup$
    – user144248
    Apr 27 '14 at 13:37
  • $\begingroup$ Probably you use $324\equiv 1\pmod{19}$. $\endgroup$
    – user144248
    Apr 27 '14 at 13:46
  • $\begingroup$ The difference between $455^{2m+1}+324$ and $455^1+324$ is equal to $455(455^{2m}-1)$ and the latter is easily seen to be divisible by $(455^2-1)$ (since $a^x-1$ is divisible by $(a-1)$). The same reasoning applies to the $2 \pmod 4$ case too. $\endgroup$ Apr 27 '14 at 18:21
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    $\begingroup$ Nice example showing that such a sequence can be proved to contain no prime without exhibiting a covering set if you can get help for some cases from an algebraic factorization. In your case an Aurifeuillean factorization. $\endgroup$ Apr 27 '19 at 8:58
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I've just checked for small $n$, and I can state that $$5 \mid 324+455^0\\17\mid 324+455^2\\89\mid324+455^4\\17\mid 324+455^6$$ I think these examples help to see a path...in fact $$324+455^{2+4k}\equiv1+455^2\cdot455^{4k}\equiv1-1\cdot(455^4)^k\equiv1-1\cdot1^k\equiv1-1\equiv0 \pmod{17}$$

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  • $\begingroup$ Thanks for these examples. But they don't yet indicate if there is an even $n$ such that $324+455^n$ is prime. $\endgroup$
    – user144248
    Apr 19 '14 at 21:01
  • $\begingroup$ I don't think this answer deserves a vote down. Of course it does not answer the question, but @Foga does not state any conclusion of that sort. $\endgroup$
    – quapka
    Apr 19 '14 at 21:04
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    $\begingroup$ The downvoter might expect Foga to post this in a comment instead of an answer. But given his edit this downvoter may change his mind... Good work Foga! $\endgroup$ Apr 19 '14 at 21:05
  • $\begingroup$ I'm working on the other case... ;) $\endgroup$
    – sirfoga
    Apr 19 '14 at 21:09
  • $\begingroup$ Just for the fun, even though the odd case is done by division by 19: $163\mid 324+455^n$ for $n \equiv 7\pmod{162}$. $29 \mid 324+455^n$ for $n \equiv 5 \pmod {28}$. $\endgroup$
    – snulty
    Apr 20 '14 at 0:14

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