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.