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Comment on Mathematicians Crack the Cursed Curveparent

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Good point, it's not necessarily possible! That said, it would be enough to know that the intersection just contains all the rational points (and is finite, though that's automatic for 2+ polynomials in two variables with no common factors). Then we can just check them one by one, discarding the non-rational points.

Alternately, it's possible the construction gives a system of auxiliary equations, which, together with f(x,y) = 0, pick out the rational points of the curve. (The term "variety", as in "Selmer variety", means solution set to a system of polynomial equations). Still, short of knowing the points in advance, I wouldn't know how to easily produce such equations.

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