TITLE: are there infinitely many triples of consecutive square-free integers?
QUESTION [17 upvotes]: The title says it all ... Obviously, any such triple must be of the form
$(4a+1,4a+2,4a+3)$ where $a$ is an integer. Has this problem
already been studied before ? The result would follow from Dickson's 
conjecture on prime patterns, which implies that there are
infinitely many integers $b$ such that $4(9b)+1,2(9b)+1$ and $4(3b)+1$ are all prime
(take $a=9b$). 
A related question : Question on consecutive integers with similar prime factorizations

REPLY [7 votes]: The question of the number of positive integers $n \leq x$ for which all members of an associated fixed pattern are squarefree (or r-free) was studied by Leon Mirsky:
L. Mirsky, Note on an asymptotic formula connected with r-free integers. 
Quart. J. Math., Oxford Ser. 18 (1947), 178-182. 
L. Mirsky, Arithmetical pattern problems relating to divisibility by rth powers. 
Proc. London Math. Soc. (2) 50 (1949), 497–508.
As I remember it, Mirsky proved that the number is $cx + O(x^{2/3})$ for patterns of squarefrees, where $c$ is a constant depending on the pattern, and is positive if the pattern is not excluded by certain necessary congruential conditions.