Three Consecutive Square-Free Integers, Infinitely Often

Created on August 01, 2026

2026   ·   mathematics

A positive integer is square-free if it is not divisible by the square of any prime. Call a positive integer $a$ simple if

\[a,\qquad a+1,\qquad a+2\]

are all square-free.

Are there infinitely many simple integers?

The answer is yes, and there is a short proof using only the density of the square-free integers.

This argument was originally written in December 2022 and published here in August 2026.

The proof

Any triple of three consecutive square-free integers must be of the form

\[4n+1,\qquad 4n+2,\qquad 4n+3.\]

Indeed, every other triple of three consecutive integers contains a multiple of $4$, which cannot be square-free.

Suppose, for contradiction, that only finitely many of these triples are entirely square-free. Then, for every sufficiently large $n$, at least one of

\[4n+1,\qquad 4n+2,\qquad 4n+3\]

is not square-free.

The integer $4n$ is also not square-free. Therefore, every sufficiently large block

\[\{4n,4n+1,4n+2,4n+3\}\]

contains at most two square-free integers.

Let $Q(x)$ denote the number of square-free positive integers not exceeding $x$. Our assumption would imply

\[Q(x)\leq \frac{x}{2}+O(1).\]

However, the square-free integers have natural density

\[\lim_{x\to\infty}\frac{Q(x)}{x} = \frac{6}{\pi^2} > \frac12.\]

This is a contradiction. Hence there are infinitely many triples of consecutive square-free integers. (\square)

A quantitative consequence

The same proof gives more than infinitude.

Let $G(N)$ be the number of $n\in{0,\ldots,N-1}$ such that

\[4n+1,\qquad 4n+2,\qquad 4n+3\]

are all square-free.

A good block contains three square-free integers, while every other block contains at most two. Consequently,

\[Q(4N)\leq 2N+G(N).\]

Since

\[Q(4N)=\frac{24}{\pi^2}N+o(N),\]

we obtain

\[\liminf_{N\to\infty}\frac{G(N)}{N} \geq \frac{24}{\pi^2}-2 \approx 0.4317.\]

Thus the proof actually establishes that such triples have positive lower density among the possible triples ((4n+1,4n+2,4n+3)).

The original note

I found this argument in December 2022 while solving Bilkent University’s Problem of the Month. The photograph below is my original handwritten solution from that time.

Handwritten proof concerning three consecutive square-free integers

Note on priority

I found this argument independently, but the result and the proof are not new. Essentially the same density argument had appeared previously, and substantially stronger asymptotic results on patterns of square-free integers are classical.

References