Furstenberg–Sárközy's theorem states that if a set of positive integers has positive upper density, then there exists (infinitely many) pair of elements of the set, whose difference is a perfect square. Since prime numbers have zero density, this theorem does not apply to them. And I was wondering if there is a result in this direction regarding prime numbers. Explicitily  are there infinitely many pairs of prime numbers $(p,q)$ such that $pq$ is a perfect square

3$\begingroup$ Almost assuredly, and that square might be less than 246. Gerhard "See About Small Prime Gaps" Paseman, 2018.10.02. $\endgroup$– Gerhard PasemanOct 2, 2018 at 17:01

$\begingroup$ Well, obvious, thank you, I will defend myself with saying that I am sleepy. However it is not proved for sure, as you say, so the question still has something actual to ask (with the problem most probably being quite easier). $\endgroup$– Stoyan ApostolovOct 2, 2018 at 17:45

1$\begingroup$ I think, should such a result be established, that this square can be $ 4 $. Googling "jumping champions" may be insightful. $\endgroup$– Sylvain JULIENOct 2, 2018 at 18:08
2 Answers
Tao and Ziegler extended the GreenTao theorem to the polynomial setting. As a very special case we get that any subset of the primes with positive relative density contains a difference which is a square.
As Joel mentions, this follows from the work of Tao and Ziegler.
Alternatively, this can be directly deduced from the density of the primes and the known bounds on the Furstenberg–Sárközy theorem. Indeed, the bounds of Pintz, Steiger, and Szemeredi from the 1980's are suffient. See:
J. Pintz, W. L. Steiger, and E. Szemeredi, On sets of natural numbers whose difference set contains no squares, J. London Math. Soc. (2) 37 (1988), 219–231
See also the even stronger bounds in the more recent work of Bloom and Maynard: https://arxiv.org/pdf/2011.13266.pdf.