Let $q$ be the next prime after $p$. Then The fact that primes exist between $(a-b)p$ which p$and$2p$implies that$a-b$must be less then$2$. This reduces the problem to finding a number$a$and prime$p$such that$ap+1, ap+2 \ldots ap+p-1$all have prime factors less then$p$while$a$has only prime factor greater then$p$. This is not a complete solution, but is a step forward. 1 Let$q$be the next prime after$p$. Then$(a-b)p$which implies that$a-b$must be less then$2$. This reduces the problem to finding a number$a$and prime$p$such that$ap+1, ap+2 \ldots ap+p-1$all have prime factors less then$p$while$a$has only prime factor greater then$p\$. This is not a complete solution, but is a step forward.