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Is there a theorem proof whose accuracy is doubted because it is short?

He told me while chatting with a friend of mine. It's about a mathematician who proves a difficult theorem very briefly and simply. Those who have examined the proof have hesitated to confirm it. "Let's not rush it because it seems too short and simple to be true!"

If this sort of thing did happen, I'm guessing it might be the proof Leroy Milton Kelly gave in 1948 for Sylvester's Line Problem (1893). Because the problem is tricky and Kelly's proof is unexpectedly short. (I have no arguments to support my guess, I am not assertive about it.)

Have you heard of a similar story? What do you think is the reality?

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    $\begingroup$ Wouldn't the period of doubt necessarily also be short, so never emerge publicly? $\endgroup$ Jan 12, 2023 at 16:11
  • $\begingroup$ @CalumGilhooley , That sounds reasonable too. $\endgroup$
    – scarface
    Jan 12, 2023 at 20:07
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    $\begingroup$ I can think of a few examples of unexpectedly short solutions to seemingly difficult problems, e.g., Lomonosov's invariant subspace theorem, Huang's proof of the sensitivity conjecture, Galvin's proof of the Dinitz conjecture, Dvir's solution of the finite field Kakeya problem. However, public acceptance of these proofs was rapid, precisely because they were so easy to check. $\endgroup$ Mar 17, 2023 at 23:05
  • $\begingroup$ So for an answer to your question, probably someone who solves such a problem has to later reveal publicly that he or she initially had private doubts about the correctness of the proof. $\endgroup$ Mar 17, 2023 at 23:07

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There's something like this in Volume II of Feller's Introduction to Probability Theory and Its Applications, Section I.5. Suppose $X_0, X_1, X_2, \ldots$ is an infinite sequence of i.i.d. real-valued random variables with a continuous distribution function. Define the waiting time $N$ as the value of the first subscript $n$ such that $X_n > X_0$. What is the expected value of $N$?

Feller gives the following argument. The event that $N > n-1$ occurs if and only if the largest term of the $n$-tuple $X_0, X_, \ldots X_{n-1}$ appears at the very beginning. By symmetry, this probability is $1/n$. Therefore, $$\Pr(N = n) = \Pr(N > n-1) - \Pr(N > n) = {1\over n} - {1\over n+1} = {1\over n(n+1)}.$$ Thus the expected value of $N$ is infinite! Feller then comments:

The striking and general nature of [this result] combined with the simplicity of the proof are apt to arouse suspicion. The argument is really impeccable (except for the informal presentation), but those who prefer to rely on brute calculation can easily verify the truth from [tedious calculation omitted].

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