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

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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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mau 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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falagar 2019-05-08 01:14:57

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