Find three consecutive pairs of integers with matching prime support (Erdős #850)

For a positive integer $m$ let $\mathrm{rad}(m)$ be its radical (the product of its distinct prime factors); two integers have the same set of prime factors iff they share the same radical. Do there exist two **distinct** positive integers $x\ne y$ such that $\mathrm{rad}(x)=\mathrm{rad}(y)$, $\mathrm{rad}(x+1)=\mathrm{rad}(y+1)$, and $\mathrm{rad}(x+2)=\mathrm{rad}(y+2)$ all hold simultaneously?

Acceptance. FULLY RESOLVES (existence direction): exhibit explicit $x\ne y$ with $\mathrm{rad}(x+i)=\mathrm{rad}(y+i)$ for $i\in\{0,1,2\}$ — trivially machine-verifiable by factoring six integers. PARTIAL: a verified exhaustive search certifying no solution with $\min(x,y)\le B$ for a stated bound $B$, or a conditional (abc) argument bounding solutions. Provide the search code / certificate.