Product of consecuative terms in subsets of primes is 1 (mod ab)

[carpetpool]

For two relatively prime integers, a, b both greater than 2, let (a, b){n} denote all the primes p which are either 1 (mod a), and or 1 (mod b), and p relatively prime to ab (n denotes the nth prime in this set, if defined). (C1)

Can it be shown that a product of all consecutive primes in (a, b){n} (to infinity) at some point will be 1 (mod ab)?

In other words, would it ever be the case that the product all primes in (a, b){n} to infinity is never 1 (mod ab) for any (a, b) pair following C1 (condition 1)?

Let P(a, b) be the smallest prime p such that the product of all primes x in (a, b){n} <= p is 1 (mod ab).


For example, P(3, 4) = 19, since the product of primes <= 19 either congruent 1 (mod 3) and or 1 (mod 4) relatively prime to 12, is 1 (mod 12).

For the first three smallest pairs, P(3, 4) = 19, P(3, 5) = 103, P(3, 7) = 283 none of which P(a, b) > 10000. The "goal" here is then to find an extrodinarily small (a, b) pair following C1, and an extrodinarily large value for P(a, b).

(3, 4)

5*7*13*17*19 = 1 (mod 12)

(3, 5)

7*11*13*19*31*37*41*43*61*67*71*73*79*97*101*103 = 1 (mod 15)

(3, 7)

13*19*29*31*37*43*61*67*71*73*79*97*103*109*113*127*139*151*157*163*181*193*197*199*211*223*229*239*241*271*277*281*283 = 1 (mod 21)

I would write a program to find P(a, b) if (a, b) follow C1 (maybe with PARI/GP), but there are too many (a, b) pairs to check. So for the purpose of this discussion, let's just stick to all satisfying C1 (a, b) pairs for a <= 30, b <= 30. (Or increase the bound if nessesary to find an extroninarily large value for P(a, b).)

here is a rough outline of the script (not generally for PARI/GP, but can) I had in mind:


(a, b){n} = list(
for k = 1, 10000
if
isprime(k*a+1) = 1,
if
gcd(k*a+1, a*b) = 1
add
list(k*a+1)

for k = 1, 10000
if
isprime(k*b+1) = 1,
if
gcd(k*b+1, a*b) = 1
add
list(k*b+1)
)

end.

P(a, n) = product((a, b){n}) for all terms <= n, = 1 (mod a*b)

end.

Any improvements, and or other working script outlines? Thanks.

[science_man_88]

Originally Posted by carpetpool

For two relatively prime integers, a, b both greater than 2, let (a, b){n} denote all the primes p which are either 1 (mod a), and or 1 (mod b), and p relatively prime to ab (n denotes the nth prime in this set, if defined). (C1)

Can it be shown that a product of all consecuative primes in (a, b){n} (to infinity) at some point will be 1 (mod ab)?

In other words, would it ever be the case that the product all primes in (a, b){n} to infinity is never 1 (mod ab) for any (a, b) pair following C1 (condition 1)?

Let P(a, b) be the smallest prime p such that the product of all primes x in (a, b){n} <= p is 1 (mod ab).


For example, P(3, 4) = 19, since the product of primes <= 19 either congruent 1 (mod 3) and or 1 (mod 4) relatively prime to 12, is 1 (mod 12).

For the first three smallest pairs, P(3, 4) = 19, P(3, 5) = 103, P(3, 7) = 283 none of which P(a, b) > 10000. The "goal" here is then to find an extrodinarily small (a, b) pair following C1, and an extrodinarily large value for P(a, b).

(3, 4)

5*7*13*17*19 = 1 (mod 12)

(3, 5)

7*11*13*19*31*37*41*43*61*67*71*73*79*97*101*103 = 1 (mod 15)

(3, 7)

13*19*29*31*37*43*61*67*71*73*79*97*103*109*113*127*139*151*157*163*181*193*197*199*211*223*229*239*241*271*277*281*283 = 1 (mod 21)

I would write a program to find P(a, b) if (a, b) follow C1 (maybe with PARI/GP), but there are too many (a, b) pairs to check. So for the purpose of this discussion, let's just stick to all satisfying C1 (a, b) pairs for a <= 30, b <= 30. (Or increase the bound if nessesary to find an extroninarily large value for P(a, b).)

here is a rough outline of the script (not generally for PARI/GP, but can) I had in mind:


(a, b){n} = list(
for k = 1, 10000
if
isprime(k*a+1) = 1,
if
gcd(k*a+1, a*b) = 1
add
list(k*a+1)

for k = 1, 10000
if
isprime(k*b+1) = 1,
if
gcd(k*b+1, a*b) = 1
add
list(k*b+1)
)

end.

P(a, n) = product((a, b){n}) for all terms <= n, = 1 (mod a*b)

end.

Any improvements, and or other working script outlines? Thanks.

like in the proof that all primes dividing 2^p-1 where p is prime are of form 2kp+1 you can note that k*a has to be even so if a is odd you need to check only even k and if a is even you can check every k ( but these two can overlap) if k has any prime divisors you can replace a with them in theory not change the number k*a+1 but test it for multiple values) gcd(k*a+1,b) is the same as gcd(k*a+1,a*b) because they're 1 and 0 mod all divisors of a respectively. same style of arguments for the other form k*b+1.