## On a research problem of C&P

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• Hello, ... 1.85. Note that 61 divides 67*71 + 1. Are there three larger consecutive primes p
Message 1 of 2 , Dec 12, 2004
Hello,

I've been looking at this research problem in Crandall & Pomerance's book:

-----
1.85. Note that 61 divides 67*71 + 1. Are there three larger consecutive
primes p < q < r such that p | qr + 1?
-----

I've also pondered the similar problem p | qr - 1, while upholding the
restrictions on p,q,r. Then it dawned on me that it implies qr - 1 == 0 mod
p, and thus qr == 1 mod p. In other words, q and r are modular inverses of
each other.

Now let q = p+s and r = p+t. Then (p+s)(p+t) == 1 mod p, so that finally we
arrive at st == 1 mod p. Now s and t are prime gaps and it is known on the
PNT that both are O(log p), so that their product is O(log^2 p). On the other
hand, st == 1 mod p implies that st = 1 + kp. This condition means that we
need, at best, st = O(p). This heuristic would suggest that no large
solutions exist (the largest I've found is 83 | 89*97 - 1).

This heuristic nicely carries over to the original statement. Now st == -1 mod
p, so that st = -1 + kp = (p-1) + (k-1)p. Again the PNT suggests st = O(log^2
p) while the congruence conditions require at least st = O(p).

Any flaws in this reasoning?

Décio
• Further info about the argument below can be found in the NMBRTHRY posting http://listserv.nodak.edu/scripts/wa.exe?A2=ind0412&L=nmbrthry&F=&S=&P=2781 There s
Message 2 of 2 , Dec 13, 2004
Further info about the argument below can be found in the NMBRTHRY posting

http://listserv.nodak.edu/scripts/wa.exe?A2=ind0412&L=nmbrthry&F=&S=&P=2781

There's a little evidence, regarding what's known about first occurrence prime
gaps, to support the conjecture.

Décio

On Monday 13 December 2004 00:52, you wrote:
> Hello,
>
> I've been looking at this research problem in Crandall & Pomerance's book:
>
> -----
> 1.85. Note that 61 divides 67*71 + 1. Are there three larger consecutive
> primes p < q < r such that p | qr + 1?
> -----
>
> I've also pondered the similar problem p | qr - 1, while upholding the
> restrictions on p,q,r. Then it dawned on me that it implies qr - 1 == 0 mod
> p, and thus qr == 1 mod p. In other words, q and r are modular inverses of
> each other.
>
> Now let q = p+s and r = p+t. Then (p+s)(p+t) == 1 mod p, so that finally we
> arrive at st == 1 mod p. Now s and t are prime gaps and it is known on the
> PNT that both are O(log p), so that their product is O(log^2 p). On the
> other hand, st == 1 mod p implies that st = 1 + kp. This condition means
> that we need, at best, st = O(p). This heuristic would suggest that no
> large solutions exist (the largest I've found is 83 | 89*97 - 1).
>
> This heuristic nicely carries over to the original statement. Now st == -1
> mod p, so that st = -1 + kp = (p-1) + (k-1)p. Again the PNT suggests st =
> O(log^2 p) while the congruence conditions require at least st = O(p).
>
> Any flaws in this reasoning?
>
> Décio

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