My question is mostly out of curiosity, with probably no other use, but here it is. I will need to provide a bit of background.
I heard from someone who works with elliptic curves that often proving that a certain curve contains no rational points at all is easier than to prove that the curve has no rational points other than a few known ones. This seems to be due to the fact that any "proof of non-existence" cannot work in full generality because it cannot be valid in those few special cases. I then thought of a similarity with a problem I studied: whether the complementary Bell numbers $b(n)$ (defined as the alternating sum of the Stirling numbers of the second kind) are ever zero except for the known case $b(2)$ =0. In that case too any attempted general proof that $b(n)$ is not zero could not work because $b(2)$ is 0.
So here is my question: I would be interested in knowing of other (maybe more interesting) examples of the fact that proving non-existence except a few special cases is especially hard when compared with a global non-existence proof in a similar setting.
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4$\begingroup$ "...any "proof of non-existence" cannot work in full generality because it cannot be valid in those few special cases." For Diophantine equations, it's more precise. Usually the first method one tries is to show that there are no solutions modulo $p$ for one (or a small handful) of primes $p$. For any given $p$, that's a finite task, and if it works, then there are also no integer solutions. But if there is an integer solution, then checking mod $p$ won't rule out integer solutions (it can't). For some equations, solutions mod $p$ for all $p$ (including $p=\infty$) implies an integer solution. $\endgroup$– Joe SilvermanCommented Oct 30 at 17:25
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$\begingroup$ In a sense any classification theorem (e.g. the classification of finite simple groups) is of this form... $\endgroup$– Sam HopkinsCommented Oct 31 at 1:22
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