Do $n=2m+1$ and $\big(2^m\bmod(m\cdot n)\big)\in\{n+1,3n-1\}$ imply $n$ prime?

Question

Equivalently, for $n=2m+1$, do $2^m\equiv\pm1\pmod n$ and $2^m\equiv2\pmod m$ imply $n$ prime?
_{Note: equivalence follows from the Chinese Remainder Theorem for $m>2$, and examination otherwise. $2^m\equiv\pm1\pmod n$ is an Euler test for $n$.}

What if we add the stronger requirements that $2^{(m-1)/2}\equiv\pm1\pmod m$ ? That $m$ pass the strong pseudoprime test to base 2?

The $n$ with $m$ prime that pass the test include all the safe primes (OEIS A005385) above $5$. The corresponding $m$ are Sophie Germain primes (OEIS A005384).
_{Proof: safe primes $p=2q+1$ match $2^q\equiv\pm1\pmod p$ by Euler's criterion, and match $2^q\equiv2\pmod q$ by Fermat's little theorem.}

I fail to prove that conversely, the $n=2m+1$ with $m$ prime that pass the test include nothing but the safe primes.

There are a few other $n$ that pass the test, dubbed pseudo-safe-primes, A300193; terms less than $2^{42}$ b300193; first ones:

683, 1123, 1291, 4931, 16963, 25603, 70667, 110491, 121403, 145771, 166667, 301703, 424843, 529547, 579883, 696323, 715523, 854467, 904103, 1112339, 1175723, 1234187, 1306667, 1444523, 2146043, 2651687, 2796203, 2882183, 3069083, 3216931, 4284283, 4325443, 4577323, 5493179, 5764531, 9949943,

The smallest even $m$ are for $n=252\,435\,584\,573$, $1\,200\,060\,997\,853$, $2\,497\,199\,739\,653$, $453\,074\,558\,824\,253$... which are prime. Any such even $m$ is an even pseudoprime (OEIS A006935).

All odd $m$ are pseudoprimes (OEIS A001567) passing a Fermat test, with the corresponding $n$ passing a strong pseudoprime test.

@Peter Košinár: that's an answer to two of out of the three questions! $n=2m+1=(2^{47}+1)/3=283*P_{12}$. Further, $m=(2^{46}-1)/3$ pass the Euler test. All that's left pending is: does there exist composite $n=2m+1$ with $n$ and $m$ passing the strong pseudoprime test? — fgrieu, Mar 06 '18 at 12:04
As you have correctly observed, I used the standard construction $m=(4^p-1)/3$ for Fermat pseudoprimes and just tested $n=(2m+1)$ for non-primality. While it guarantees $m$ to satisfy Euler criterion too, it fails the extended Euler-Jacobi version of it where $\pm 1$ is replaced by Jacobi symbol (and thus also fails the strong pseudoprimality test). When I tried to do something similar with one specific explicit construction of strong pseudoprimes, it failed miserably: the resulting $n$ failed even the Fermat test. I will keep playing with a few more ideas, though. — Peter Košinár, Mar 06 '18 at 13:02

fgrieu · Answer 1 · 2018-03-18T20:04:22.127

After Peter Košinár's answer/comment, it is settled that the answer to the title's question is no. And that even the stronger requirements: $n=2m+1$, $2^m\equiv\pm1\pmod n$, and $2^{(m-1)/2}\equiv\pm1\pmod m\ $ do not imply $n$ prime. Many of the $m=(4^p-1)/3$ with $p$ prime turn out to be counterexamples, including $p$ among

23, 29, 41, 53, 89, 113, 131, 179, 191, 233, 239, 251, 281, 293, 359, 419, 431, 443, 491, 509, 593, 641, 653, 659, 683, 719, 743, 761, 809, 911, 953, 1013, 1019, 1031, 1049, 1103, 1223, 1229, 1289, 1409, 1439, 1451, 1481, 1499, 1511, 1559, 1583, 1601, 1733, 1811, 1889, 1901, 1931, 1973, 2003,

It holds that if $n=2m+1$ with $m$ prime and $2^m\equiv\pm1\pmod n$, then $n$ is prime. That's proven by Fedor Petrov here, with details there.

Question remaining open: do $n=2m+1$, $n$ and $m$ passing the strong pseudoprime test to base 2 imply $n$ prime?

Do $n=2m+1$ and $\big(2^m\bmod(m\cdot n)\big)\in\{n+1,3n-1\}$ imply $n$ prime?

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