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.
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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. |
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