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I think conjecture 3 is actually stronger then conjecture 4.

I prove $C_3\implies C_4$:

Pick any sequence of integers $a_n$, which contains all integers infinite times.

Pick any enumeration of all squares $s_n$ in the plane with corners at rational coordinates.

Then assuming conjecture 3, at step $n$ we can find a point with rational distances in $s_{a_n}$ which is not collinear to any previously used rationals, since there are only finitely many straight lines in our set so far.

At step $\omega$, we have $\omega$ many rationals in every square, so a dense uniform set of all rational distances.

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Doesnt

I think conjecture 3 is actually imply stronger then conjecture 4?.

I prove $C_3\implies C_4$:

Pick your favorite any sequence of integers $a_n$, which contains all integers infinite times.

Pick any enumeration of all squares $s_n$ in the plane with corners at rational coordinates.

Then assuming conjecture 3, at step $n$ we can find a point with rational distances in $s_{a_n}$ which is not collinear to any previously used rationals, since there are only finitely many straight lines in our set so far.

At step $\omega$, we have $\omega$ many rationals in every square, so a dense uniform set of all rational distances.

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Doesnt conjecture 3 actually imply conjecture 4?

Pick your favorite sequence of integers $a_n$, which contains all integers infinite times.

Pick any enumeration of all squares $s_n$ in the plane with rational coordinates.

Then assuming conjecture 3, at step $n$ we can find a point with rational distances in $s_{a_n}$ which is not collinear to any previously used rationals, since there are only finitely many straight lines in our set so far.

At step $\omega$, we have $\omega$ many rationals in every square, so a dense uniform set of all rational distances.