A classic theorem of Fermat states that there are no four squares in arithmetic progression.

Without the coprimality condition this is easy, since if $a,a+d,\ldots,a+(k-1)d$ is an arithmetic progression of powerful numbers then so too is\[a(a+kd)^2,\ldots,(a+(k-1)d)(a+kd)^2,(a+kd)^3.\]Beginning with $k=2$ and an arbitrary pair of powerful numbers one can construct arbitrarily long arithmetic progressions of powerful numbers.

One can similarly ask for coprime arithmetic progressions in the $r$-powerful numbers (i.e. if $p\mid n$ then $p^r\mid n$). Erdős [Er76d] conjectured that when $r\geq 4$ there do not exist infinitely many such progressions of length $3$, and when $r=3$ there are infinitely many progressions of length $3$ but only finitely many of length $4$.

Bajpai, Bennett, and Chan [BBC24] proved that there are infinitely many four-term progressions of coprime powerful numbers, and infinitely many three-term progressions of coprime $3$-powerful numbers. They also show that there exist only finitely many three-term coprime progressions when $r\geq 4$ assuming the ABC conjecture.

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This page was last edited 31 October 2025.