There are primes $p,q$ such that $\gcd(a,b)=|ap-bq|$

Question

Conjecture:

Given $a,b\in\mathbb Z^+$ there are primes $p,q$ such that $\gcd(a,b)=|ap-bq|$.

I would like help with a proof or a counter-example. Tested for millions of pseudo random numbers.

Answer 1

Let $a, b>2$ with $b$ very larger than $a$. Suppose the twin prime conjecture true and $b-a=2$

be prime numbers, $gcd(a, b) =1=pa-qb$. If $p, q$ are prime, $ap$ and $qb$ are odd, and $ap-qb$ is even. Contradiction. You cannot have $p=2$ or $q=2$ if $a, b$ are enough big. Suppose, $a, a+2$ are prime, and $q=2$. $p$ is odd so at least $3$. $3a-2(a+2)=a-4$ if $a$ is not big it is not possible.