are there infinitely many triples of consecutive square-free integers? 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
 A: 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.
A: To expand on the answer in my comment, the proportion of integers $a$ for which $4a+1,4a+2,4a+3$ are all squarefree is $\prod_{p\not=2}(1-3/p^2)$, with the product taken over all odd primes $p$. As this product converges to a positive limit, there are infinitely many such $a$. A quick heuristic is to look modulo $p^2$. For odd prime $p$, precisely $p^2-3$ of the possible $p^2$ values of $a$ mod $p^2$ lead to $4a+1,4a+2,4a+3$ all being nonzero mod $p^2$, so this has probability $1-3/p^2$. Independence of mod $p^2$ arithmetic as $p$ runs through the primes suggests the claimed limit.
More precisely, if $\phi(n)$ is the number of positive integers $a\le n$ with $4a+1,4a+2,4a+3$ squarefree then
$$
\frac{\phi(n)}{n}\to\prod_{p\not=2}(1-3/p^2).\qquad\qquad{\rm(1)}
$$
It's not too hard to turn this heuristic into a rigorous argument. If we let $\phi_N(n)$ denote the number of $a\le n$ such that none of $4a+1,4a+2,4a+3$ is a multiple of $p^2$ for a prime $p < N$, then the Chinese remainder theorem says that we get equality
$$
\frac{\phi_N(n)}{n}=\prod_{\substack{p\not=2,\\ p < N}}(1-3/p^2).\qquad\qquad{\rm(2)}
$$
wherever $n$ is a multiple of $\prod_{\substack{p\not=2,\\ p < N}}p^2$ and, therefore, the error in (2) is of order $1/n$ for arbitrary $n$. It only needs to be shown that ignoring primes $p\ge N$ leads to an error which is vanishingly small as $N$ is made large. In fact, the number of $a \le n$ which are a multiple of $p^2$ is $\left\lfloor\frac{n}{p^2}\right\rfloor\le \frac{n}{p^2}$. The number of $a\le n$ which is a multiple of $p^2$ for some prime $p\ge N$ is bounded by $n\sum_{p\ge N}p^{-2}$. So, the proportion of $a\le n$ for which one of $4a+1,4a+2,4a+3$ is a multiple of $p^2$ for an odd prime $p\ge N$ is bounded by $3\frac{4n+3}{n}\sum_{p\ge N}p^{-2}\sim3(4+3/n)/(N\log N)$. This means that $\phi_N(n)/n\to\phi(n)/n$ uniformly in $n$ as $N\to\infty$, and the limit (1) follows from approximating by $\phi_N$.
A: I found these answers by Erick Wong. The simpler version is that if the answer was no, then at least two of every $4a,4a+1,4a+2,4a+3$ must not be squarefree for $a$ large enough, so the density of squarefree numbers would be limited by $1/2$. But it is $6/\pi^2>1/2$ so there must be infinitely many consecutive triples of squarefree numbers.
By the way, I don't think Dickson's conjecture applies, since one of $4p+1,2p+1,4p+3$ is divisible by 3.
A: The answer is yes. More precisely, George Lowther's heuristic is right, i.e. the density of $n$'s such that $n-1$, $n$, $n+1$ are square-free is $\prod_p(1-3/p^2)$ over all primes $p$. 
To see this, let $P$ be fixed but large. As $x\to\infty$, the number of $n\leq x$ such that none of $n-1$, $n$, $n+1$ is divisible by the square of some prime $p\leq P$ is $x\prod_{p \leq P}(1-3/p^2)+o(x)$ by the Chinese Remainder Theorem. The number of $n\leq x$ such that one of $n-1$, $n$, $n+1$ is divisible by the square of some prime $p>P$ is at most $\sum_{P < p\leq \sqrt{x+1}} 3\left(\frac{x}{p^2}+1\right)\ll\frac{x}{P}+O(\sqrt{x})$. Hence the number of $n\leq x$ in question is $x\prod_{p \leq P}(1-3/p^2)+O\left(\frac{x}{P}\right)+o(x)$. Letting $P\to\infty$ proves the claim.
A more careful count, e.g. the choice $P:=(\log x)/10$, reveals that the number of $n\leq x$ in question is $x\prod_p(1-3/p^2) + O(x/\log x)$.
BTW these questions have a large literature. See e.g. some of Harald Helfgott's papers with the words "square-free sieve" or "power-free values" in the title, and the references in them.
