Ergodic variant of the quadratic Rado conjecture

About 1 year old · traced to

Let (a,b,c)(a,b,c) be a Rado triple, meaning that a=ca=c, b=cb=c, or a+b=ca+b=c. A multiplicative action is a quadruple (X,X,μ,Tn)(X,\mathcal{X},\mu,T_n) where (X,X,μ)(X,\mathcal{X},\mu) is a probability space and the invertible measure-preserving transformations TnT_n satisfy T1=id⁡T_1=\operatorname{id} and Tm∘Tn=TmnT_m\circ T_n=T_{mn}. Let A∈XA\in\mathcal{X} satisfy

X=⋃j=1kTj−1AX=\bigcup_{j=1}^kT_j^{-1}A

for some k∈Nk\in\mathbb{N}.

Ergodic variant of the quadratic Rado conjecture. There exist distinct x,y,z∈Nx,y,z\in\mathbb{N} such that

μ(Tx−1A∩Ty−1A∩Tz−1A)>0\mu\bigl(T_x^{-1}A\cap T_y^{-1}A\cap T_z^{-1}A\bigr)>0

and

ax2+by2=cz2.a x^2+b y^2=c z^2.

This multiple-recurrence statement is presented as an ergodic-theoretic reduction of the quadratic Rado conjecture. The cited results establish related pair configurations and special coefficient cases, but the stated triple recurrence problem remains open.

References

Primary source

Nikos Frantzikinakis and Andreas Mountakis, “Recurrence for pretentious systems along generalized Pythagorean triples”, arXiv:2508.09778 (2025).

Progress summary

Never refreshed

Nothing recorded yet. Refresh searches the literature and the public web for attempts on this problem, and writes the first summary here.

Solutions 0

No solutions have been posted yet.