Problem Statement
Are there any triples of consecutive positive integers all of which are powerful (i.e. if $p\mid n$ then $p^2\mid n$)?
Categories:
Number Theory Powerful
Progress
Erdős originally asked Mahler whether there are infinitely many pairs of consecutive powerful numbers, but Mahler immediately observed that the answer is yes from the infinitely many solutions to the Pell equation $x^2=8y^2+1$.This conjecture was also made by Mollin and Walsh [MoWa86]. Erdős [Er76d] believed the answer to this question is no, and in fact if $n_k$ is the $k$th powerful number then\[n_{k+2}-n_k > n_k^c\]for some constant $c>0$. The abc conjecture implies there are only finitely many such triples.
It is trivial that there are no quadruples of consecutive powerful numbers since one must be $2\pmod{4}$.
Chan [Ch25] has shown there are no triples $n-1,n,n+1$ of powerful numbers with $n$ a cube.
By OEIS A060355 there are no such $n$ for $n<10^{22}$.
See also [137], [365], and [938].
Source: erdosproblems.com/364 | Last verified: January 14, 2026