Suppose $p$ and $q$ are prime numbers. Let’s call the pair $(p, q)$ a Fermat pair iff $| p – q | < 2\sqrt{2} (pq)^{\frac{1}{4}}$.
Such prime pairs possess a rather interesting property: they can neatly be expressed using only their product (even though prime decomposition is generally a computationally hard problem). If I recall correctly, this fact was first discovered by Pierre de Fermat (hence the name of the pairs).
To be concrete, if $N = pq$ and $p > q$, then
$$p = \lfloor \sqrt{N} \rfloor + 1 + \sqrt{(\lfloor \sqrt{N} \rfloor + 1)^2 – N}$$
$$q = \lfloor \sqrt{N} \rfloor + 1 – \sqrt{(\lfloor \sqrt{N} \rfloor + 1)^2 – N}$$
Proof:
Suppose $a = \frac{p + q}{2}$ and $b = \frac{p – q}{2}$. Then $N = a^2 – b^2$. Now, from $b < \sqrt{2} N^{\frac{1}{4}} < \sqrt{2a}$ we can derive that $(a-1)^2 \leq N < a^2$, which yields us our formulas.
My question is:
Are there infinitely many Fermat pairs?
If the answer is known, it should be positive. Why? Because any pair of twin primes is a Fermat pair. Therefore the negative answer to this question would have given negative answer to the Twin Prime Conjecture (and that problem is currently open).
However, if it is known, I would like to see a proof that there are infinitely many Fermat pairs.
Best Answer
It has been proven that there an infinitely many primes that differ by at most 246. Therefore, with $p > q$, you have infinitely many cases such that $p - q \leq 246$.
But, solving $2\sqrt{2}(pq)^{\frac{1}{4}} > 246$ yields $pq > \left(\frac{123}{\sqrt{2}}\right)^4$. Since $p = q + k$ with $k$ at most $246$, this can readily be solved to show that there is a finite lower bound on $p$, for which all $p$ bigger than this bound solves the inequality thus satisfying the condition of a Fermat pair.
Links can be found here: https://asone.ai/polymath/index.php?title=Bounded_gaps_between_primes