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

  • 2
    Surely the proportion of integers $a$ satisfying this property is $\prod_p(1-3/p^2)$, with the product over odd primes $p$? So, yes, there are infinitely many. – George Lowther Mar 27 '11 at 17:35
  • 2
    @Qiaochu: The probability of none of $4a+1,4a+2,4a+3$ being a multiple of $p^2$ is $1−3/p^2$, for odd primes $p$ (and it is 1 for $p=2$). Treating these as independent events, you multiply the probabilities together. This is just a heuristic, of course, but I don't think it's hard to turn into a rigorous proof. A triple of consecutive integers has zero probability of all being even, just looking at it mod 2, so the same kind of heuristic gives the proportion of triples of consecutive even integers to be zero, as you'd hope. – George Lowther Mar 27 '11 at 18:15
  • 1
    It might be added to George's comments that you can truncate his product to $p < Y$ with error roughly on the order of $1/Y$ (imagine you are looking for triples in $[X, 2X]$ and take, say, $Y = X^{1/10}$). This is why his proof works here but you can't prove, say, the Twin Prime Conjecture using a similar argument. – Frank Thorne Mar 27 '11 at 18:42
  • 2
    A lot of the comments seem to invoke unnecessary sieve theory. Does this simpler argument work? Look at the integers that are not 0 mod 4. Among these, easy argument gives $8/\pi^2$ squarefree, or 19% with a nontrivial square factor. As 19% < 1/3, the proposition follows by pigeonholes, with some density. For more precision, extra work is needed. – Junkie Mar 27 '11 at 19:41
  • 4
    @GH: I guess I don't understand why it is difficult to get something good enough. Only 1/9 of the numbers are divisible by $3^2$, and so on, with an error (remembering mod 4) of no more than 4. Don't bother with fancy inclusion and exclusion, and directly sum $\sum_{n<\sqrt{X}} X/n^2+O(4)$ for $n>1$ odd as an upper bound for counting of those divisible by odd squares. The sum is easily bounded by $0.24X$ without much work, so is less than 1/3. The upper bound here is not close to $1−8/\pi^2$ but that is not needed – Junkie Mar 27 '11 at 23:07
up vote 28 down vote accepted

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 $\phi_N$.

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.

  • @ Zander : Erick Wong's argument is correct, but actually it's an unnecessarily involved variant of George Lowther's, and it does not yield the density of the set of solutions (the product of 1-(3/(p^2))). – Ewan Delanoy Mar 27 '11 at 18:14
  • @Zander : I corrected the Dickson conjecture part, thanks. – Ewan Delanoy Mar 27 '11 at 18:15
  • 1
    Unnecessarily involved?? De gustibus non est dispudandum... – user9072 Mar 27 '11 at 18:28
  • Well, "disputandum" – Yemon Choi Apr 1 '11 at 23:01
  • Dickson's conjecture does apply, you just have to use (say) 12n+1, 6n+1, and 4n+1. But that's a pretty big sledgehammer to apply to a little problem like this! – Charles Mar 5 '13 at 14:51

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.

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.

Your Answer


By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.