Conjecture: There are infinitely many $N$ such that $0\equiv N\pmod2$, $1\equiv N\pmod3$, and $0\leq N\pmod P\leq P-4$ for primes $P$ with $5\leq P<N$

Question

Here's a conjecture,

There are infinitely many numbers $N$, such that for all prime numbers $P<N$

$0 \equiv N \pmod 2$ and
$1 \equiv N \pmod 3$ and
$0 \leq N \pmod P \leq (P-4)$, for $P \geq 5$

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For eg: 10

$10$ mod $2$ $\equiv 0$
$10$ mod $3$ $\equiv 1$
$10$ mod $5$ $\equiv 0$
$10$ mod $7$ $\equiv 3$

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Is this a known conjecture? Does it have a proof?

If not, is there a way to prove or disprove this?

Answer 1

The first two conditions force N = 4 (modulo 6) and N+2 is a multiple of 6.

If N + 2 = 6pq where p >= 5 is a prime, then N+2 is divisible by p, therefore N = (p-2) modulo p, so N doesn’t meet the conditions. Therefore the only choices for N are $N = 2^m 3^n - 2$ where n, m >= 1.

There are 1178 such numbers $N < 2^{64}-2$. 62 don’t fail for any prime < 1,000. 32 don’t fail for any prime < 100,000 and 27 don’t fail for any prime p.

The largest N is $N = 2^{32} 3^{15} - 2<2^{56}$. Numerical results are inconclusive either way.

For the curious:

2^ 1 * 3^ 1 - 2 =                      4
2^ 2 * 3^ 1 - 2 =                     10
2^ 1 * 3^ 2 - 2 =                     16
2^ 3 * 3^ 2 - 2 =                     70
2^ 2 * 3^ 3 - 2 =                    106
2^ 6 * 3^ 1 - 2 =                    190
2^ 4 * 3^ 3 - 2 =                    430
2^ 7 * 3^ 2 - 2 =                  1,150
2^ 5 * 3^ 4 - 2 =                  2,590
2^ 6 * 3^ 7 - 2 =                139,966
2^ 3 * 3^10 - 2 =                472,390
2^18 * 3^ 1 - 2 =                786,430
2^12 * 3^ 5 - 2 =                995,326
2^ 2 * 3^15 - 2 =             57,395,626
2^18 * 3^ 5 - 2 =             63,700,990
2^21 * 3^ 4 - 2 =            169,869,310
2^24 * 3^ 5 - 2 =          4,076,863,486
2^27 * 3^ 4 - 2 =         10,871,635,966
2^30 * 3^ 7 - 2 =      2,348,273,369,086
2^33 * 3^ 8 - 2 =     56,358,560,858,110
2^43 * 3^ 2 - 2 =     79,164,837,199,870
2^32 * 3^ 9 - 2 =     84,537,841,287,166
2^36 * 3^ 7 - 2 =    150,289,495,621,630
2^11 * 3^24 - 2 =    578,415,690,713,086
2^31 * 3^12 - 2 =  1,141,260,857,376,766
2^43 * 3^ 8 - 2 = 57,711,166,318,706,686
2^32 * 3^15 - 2 = 61,628,086,298,345,470

Answer 2

I changed the question slightly to an equivalent question: Are there infinitely many N = 4 modulo 6, so that N+1, N+2 and N+3 are not divisibly by any prime 5 <= p < N?

N+2 must be of the form $2^n 3^m$: If N+2 = 6k is not of this form, then k has a prime factor p < N. So we restrict N to the form $N = 2^n 3^m - 2$, and since N+2 has no prime factor ≥ 5, the question is: Does N+1 or N+3 have such a prime factor?

N+1 = 5 modulo 6, and N+3 = 7 modulo 6. N+1 and N+3 are both greater than N, so the question is: Is either N+1 or N+3 composite? So solutions are $N = 2^n 3^m - 2$ where N+1 and N+3 are twin primes.

For random x, the chance that (x, x+2) are twin primes is about $0.66 / log^2 x$. But with x = 5 (modulo 6), neither is divisible by 2 or 3, where for the other cases at least one is divisible by 2 or 3, so the chances are 6 times hight, about $4 / log^2 x$. There are about $n / log_2(3)$ candidates $2^{n-1} < N < 2^n$, each is a solution with probability $4 / n^2$, so I expect about $4 / log_2(3) / n$ solutions between $2^{n-1} < N < 2^n$. That's the harmonic series, so about $4 log n / log_2(3)$ solutions N < 2^n. This would be a very, very slowly growing function.

So this heuristic seems to show a very, very slowly growing function going towards infinity. But since its growing so slowly, any heuristics that reduces the number just a tiny bit may be enough to make this false.

To find examples for N: Generate numbers $N = 2^n 3^m - 2$, n, m ≥ 1, then check whether N+1 or N+2 have small prime factors, and if need perform a non-deterministic primality test for N+1 and N+3.