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Two relatively prime integers, $a$ $>$ $2$ and $b$ $>$ $2$, let $(a, b)[{n}]$ denote all primes $p$ which are either $1$ $\mod a$, and or $1$ $\mod b$, and $p$ relatively prime to ab ($n$ denotes the $n$th prime in this set, if defined).

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

In other words, would it ever be the case that the product all primes in $(a, b)[{n}]$ to infinity is never $1$ $\pmod {ab}$.

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

For example, $P(3, 4) = 19$, since the product of primes $\le 19$ either congruent $1$ $\pmod 3$ and or $1$ $\pmod 4$ relatively prime to $12$, is $1$ $\pmod {12}$.

The first three smallest pairs, $P(3, 4) = 19$, $P(3, 5) = 103$, $P(3, 7) = 283$.

$(3, 4)$

$5*7*13*17*19$ $=$ 1 $\pmod {12}$

$(3, 5)$

$7*11*13*19*31*37*41*43*61*67*71*73*79*97*101*103$ $=$ $1$ $\pmod {15}$

$(3, 7)$

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

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    The formulation "either...and or" is unlucky. Why not just use "or" ? This includes the possibility that both conditions are met. – 2017-03-02
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    A proof will be probably very difficult, but the results of my program after painful long calculation (without a light on the horizon) gives some evidence that the conjecture might actually be true. Of course, the limit I have reached is not very high, but at least for the pairs upto $(99/100)$, the conjecture is true. – 2017-03-02
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    I finished the verification for $1\le a$200$, we always reach $1$ – 2017-03-02

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Not an answer, but I programmed a function $q(a,b)$ in PARI/GP and I checked the coprime pairs ($a,b$) with $1\le a

It is $16403311$.

I will continue the search.

I do not really have an idea how to prove that we always eventually get the desired equivalence, but since $(a,b)$ are coprime the chinese remainder theorem might help.

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    For the pair $(73,199)$, we have to go until prime $123947947$. I stopped the program because it needed much time, so the range upto pair $(199,200)$ is not finished yet. – 2017-03-02
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    For the pair $(149/197)$, we have to go until prime $203578619$. My search is currently at $a=161$ (lexicographically ordered), so the range upto $(199/200)$ will not take too long anymore. But the specialist for such tasks , Dana Jacobsen , will surely find an implementation running much faster! – 2017-03-02