Nonprimes with $3^{n-1} \equiv 2^{n-1} \pmod n$

Is it real that there are definitely several nonprime integers $n$ such that $3^{n-1} - 2^{n-1}$ is a numerous of $n$?

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2019-05-07 13:32:54
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Answers: 2

as Qiaochu Yuan mentioned, take a Carmichael number q ; necessarily, 3 q - 1 and also 2 q - 1 are both conforming to 1 mod q , so their distinction is a numerous of q . Given that Carmichael numbers are boundless, you are done.

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2019-05-09 08:59:00
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I assume I recognize just how to confirm it without use Carmichael numbers.

Take $n = 3^{2^k} - 2^{2^k}$. After that it's not tough to confirm by induction on $k$ that $2^k | n - 1$. Therefore $3^{2^k} - 2^{2^k} | 3^{n-1} - 2^{n-1}$ (that's real due to the fact that $a - b | a^r - b^r$ for any kind of all-natural $a, b$ and also $k$).

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2019-05-08 01:14:57
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