Feedback on a general conjecture for A = Bp – Cq where p and q are primes?

I previously asked for feedback on the following conjectures (similar in appearance to Goldbach ‘s and Lemoine’s see: Two new conjectures related to Lemoine's and Goldbach's) : $O = 2n + 1 = p - 2q$ always has a solution in distinct odd primes $p$ and $q$ for $O \ge 5$ $O = 2n + 1 = 2q - p$ always has a solution in $p$ and $q$ with $q < p$, $p$ being an odd prime and $q$ being 1 or an odd prime and $O \ge 1$

Those two conjectures, even though they do not seem easy to prove, are very weak: the number of possible solutions is possibly infinite. (Since increases in $p$ and $q$ have opposite effect.) Also, while it is possible by concrete calculations to increase the domain of proven validity, it is not possible to find a concrete counter example since a solution may lay with higher values of $p$ and $q$.

Since they are so easy to satisfy, I thought that they may become more interesting if they were generalized by changing the component $2q$ into $2nq$. It is easy to see that in $O = p-2nq$ we will not have solutions in primes $p$ and $q$ as long as $n$ and $O$ have common divisors. So the revised conjectures become:

For $O$, odd natural number, and $n$ ,natural number ,without common divisors: $O = p – 2nq$ always has a solution in distinct odd primes $p$ and $q$ for $o \ge 5$ $O = 2nq – p$ always has a solution in $p$ and $q$ with $q < p$, $p$ being an odd prime and $q$ being 1 or an odd prime and $o \ge 1$

I verified that it is the case for all $O < 2000000$ and $n < 1000$

But better, a further generalization is possible that envelops these conjectures:

$A = Bp – Cq$ where the natural numbers $A,B,C$ have no common divisor and one of them is even has solutions in $p$ and $q$ odd primes or 1

I verified this with values of $A, B$ and $C$ ranging from 1 to 1000.

This is not an answer, just a trivial observation.

If there are infinitely many twin primes then if $O=p-2q$ can be represented in that way then so can $O+2$, if $p$ and $p+2$ are twin primes. The same for a second "$O$-conjecture" for $O-2$ if $p$ and $p-2$ are twin primes.

This can be made arbitrarily long by Green-Tao theorem, for some $p$´s

If $O=p-2q$ then $O+2q=p=O+q+q$ so maybe you will try to improve some results of Vinogradov to see when it can happen that in summative decomposition some prime can appear two times (when $O$ is itself prime). Similarly for $O=2q-p$ (when $O+2p$ is prime).

$O=2q-p$ has some flavour of weak Goldbach because of $O+2p=q+q+p$ so you want (?) to know when it can happen that in decomposition into a sum there must be two same primes.

But I think that de Polignac could be the key.

• Thanks! I am not a mathematician but your "trivial observations" are interesting. – Raymond Duchesne Oct 2 '17 at 21:26
• Neither am I. @RaymondDuchesne – user480281 Oct 2 '17 at 21:28
• I was told that this is a special case of Dickson's Conjecture. ( A = Bx-Cy can be written as x = a1+b1n, y = a2+b2n for some integers a1, a2, b1 and b2 depending on A,B and C.) It seems that if we try to reconcile my conjecture with the two "Dicksonian equations", we can with: B = b1, C = b2 and A = a2 * B – a1 * C ... – Raymond Duchesne Oct 5 '17 at 20:29
• Following on my previous comment: X = a1 + B * n and Y = a2 + C * n (B & C) are co-primes ; it is easy to see that a1 and B are also co-primes and so are a2 and C. Dirichlet’s theorem tells us that x = a1 + B * n will accept an infinite number of solutions with x prime. So will y = a2 + C * n for y prime. For changing n, x and y would take an infinite number of prime values; can we prove that they necessarily sometimes have prime values for the same n? The increase intervals B and C are co-primes. – Raymond Duchesne Oct 5 '17 at 20:45