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Oct 19, 2021 at 16:45 comment added Daniel Loughran @WillSawin Ha this is a very funny trick. Why not add it as an answer?
Oct 19, 2021 at 15:00 comment added Will Sawin @KConrad In fact, shouldn't we be able to make a rigorous such implication in this case by taking a family and base changing it along a map $\mathbb P^1 \to \mathbb P^1$ whose induced map $\mathbb P^1(\mathbb Q) \to \mathbb P^1(\mathbb Q)$ misses any given finite set of points?
Oct 19, 2021 at 14:13 vote accept Jonathan Love
Oct 17, 2021 at 20:10 comment added KConrad Sometimes finding infinitely many examples in an individual case is subtle. For example, it's easy to show there are infinitely many primes $p \equiv 1 \bmod 5$ using the 5th cyclotomic polynomial and a modification of Euclid's proof of infinitude of the primes, but I've never seen a proof that there are infinitely many primes $p \equiv 2 \bmod 5$ that is not essentially like the general proof of Dirichlet's theorem on primes in arithmetic progression. Of course it's pretty easy to see there is at least one prime $p \equiv 2 \bmod 5$ (or at least two primes, or at least three primes, or ...).
Oct 17, 2021 at 20:07 comment added KConrad In the spirit of "as hard to show one as to show infinitely many", see mathoverflow.net/questions/383580/… and mathoverflow.net/questions/226794/…. The key point I am making is that while it is often easy in practice to find one example in individual cases, trying to prove there is one example in all cases is where proofs appear to be or are really known to be basically just as hard as proving there are infinitely many examples in all cases.
Oct 15, 2021 at 11:01 history undeleted Daniel Loughran
Oct 15, 2021 at 11:01 history deleted Daniel Loughran via Vote
Oct 15, 2021 at 11:01 history answered Daniel Loughran CC BY-SA 4.0