Proving that there is an infinite number of pairs of prime numbers for which $F(n)F(n+1) =pq $ for no $n>1 \in \mathbb{N}$, $F(n)$ is the GPF function

Question

Proving that there is an infinite number of pairs of prime numbers for which $F(n)F(n+1) = pq $ does not hold for any $n>1 \in \mathbb{N}$, $F(n)$ is the GPF function

I have been trying to solve this problem involving the greatest prime factor function (A006530). Two primes hold the desired property if their product is equal to the product of the GPF of two consecutive natural numbers. A brief look at the function will reveal that many of the small primes will indeed work, but we are supposed to get an infinite number of pairs which cannot have the property.

How to approach this problem? OEIS does not provide any useful piece of information regarding the function itself, so perhaps something connected with prime distributions directly?

Thank you in advance, looking forward to your insights

Answer 1

For any odd prime number $p$, the pair $(p, F(p+1))$ has the desired property. As $p > F(p+1)$, these pairs are all distinct. Given the infinity of odd prime numbers, the statement is proven.

Answer 2

Constructing a pair: Let $p$ be an odd prime such that $2^p-1$ is composite. Then $2^p-1$ is not a prime power by Mihăilescu's theorem, so we can define $q$ as its second largest prime factor. I claim that there do not exist integers $n>1$ with $F(n)F(n+1)=2q$.

Proof: By contradiction. Assume such an $n$ exists. We consider two cases. First, if $F(n)=2$, then $n=2^m$ for some integer $m$, and $2^m\equiv -1\pmod q$. However, $2$ has order $p$ in $\mathbb{F}_q^\times$ and $-1$ has order $2$. This gives a contradiction, since $q$ is odd.

Next, $F(n+1)=2$. Then $n=2^m-1$ for some positive integer $m$, so $2^m\equiv 1\pmod q$. Because the period of $2$ in $\mathbb{F}_q^\times$ is $p$, it follows that $p\mid m$, whence $2^p-1\mid 2^m-1$, whence the largest prime factor of $2^p-1$ (which, by construction, is larger than $q$) divides $2^m-1$. Again, we have a contradiction.

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Does this method give infinitely many pairs? It's easy to see that each 'seed prime' $p$ gives a different prime $q$, since if $q\mid 2^{p_1}-1,2^{p_2}-1$, then $2$ would have order both $p_1$ and $p_2$ in $\mathbb{F}_q^{\times}$, which is absurd when $p_1\neq p_2$.

Therefore this method yields infinitely many pairs if and only if there exist infinitely many primes $p$ such that $2^p-1$ is composite. Mersenne Primes are extremely rare, so there is overwhelming evidence, but I'm not (yet) aware of a proof.