majer

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2026-10-06 06:56:05 UTC

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About “they lie on a perfectly straight line is zero”，here is a slightly overengineered proof.
Write the three points as
\( P_i=(X_i,Y_i) \)
The sample space is
\[ D^3\subset\mathbb R^6 \]
where \(D\) is the unit disk.
The three points are collinear if and only if
\[\begin{vmatrix} X_1&Y_1&1\\ X_2&Y_2&1\\ X_3&Y_3&1 \end{vmatrix}=0 \]
In other words,
\[ f(X_1,Y_1,X_2,Y_2,X_3,Y_3)=0 \]
where \(f\) is a polynomial in six variables.
Thus the set of collinear triples is
\(A=f^{-1}(0) \)
an algebraic hypersurface in \(\mathbb R^6\). Its dimension is at most \(5\), and hence it has \(6\)-dimensional Lebesgue measure zero.
On the other hand, the joint distribution of the three independently and uniformly chosen points is absolutely continuous with respect to \(6\)-dimensional Lebesgue measure.
Therefore,
P(A)=0

2026-07-23 09:36:26 UTC

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Replace 2 with 1662, then 𝑝 is a Wieferich prime to base 1662. Earlier this year, for the first time, a prime 𝑝 was discovered satisfying
\( 1662^{p-1} \equiv 1 \pmod{p^2} \) .
In this case,𝑝=4001605247257.
Previously, no prime below 4 trillion had been found to satisfy this congruence. The discoverer was Richard Fischer.https://www.fermatquotient.com/FermatQuotienten/FermQ_Sorg.txt
But there is deeper mathematics here: it is not merely the discovery of a curious prime. From this, one can deduce that there exists a natural number 𝑛 such that the ring of integers of \(\mathbb{Q}(\sqrt[n]{1662})\) is not \( \mathbb{Z}[\sqrt[n]{1662}] \) . The smallest such example is precisely 𝑛=4001605247257.
In the literature (https://kconrad.math.uconn.edu/blurbs/gradnumthy/integersradical.pdfsee Theorem 5.3), it is stated:
When 𝑎≠±1,𝑛≥2, and \( T^n - a\) is irreducible over 𝑍,the ring of integers of \(\mathbb{Q}(\sqrt[n]{a})\) equals \( \mathbb{Z}[\sqrt[n]{a}] \) if and only if 𝑎 is square-free, and none of the prime divisors of 𝑛 is a Wieferich prime to base 𝑎.
Thus, when 𝑎=1662, the polynomial \( 𝑇^𝑛−1662 \) satisfies:−1662 is divisible by 2 but not by 4,
all middle coefficients are 0, hence divisible by 2,the leading coefficient (=1) is not divisible by 2.
Therefore, by Eisenstein’s criterion with 𝑝=2,\( 𝑇^𝑛−1662 \) is irreducible over 𝑄.
Since 1662 is square-free, for all 𝑛 less than the Wieferich prime to base 1662 (namely
4001605247257), the ring of integers is \( \mathbb{Z}[\sqrt[n]{1662}] \) .
Similarly, for \(\mathbb{Q}(\sqrt[n]{2})\) , when 𝑛=1093, the ring of integers is not \( \mathbb{Z}[\sqrt[n]{2}] \) , because 1093 is the classical Wieferich prime to base 2.
https://media.mathstodon.xyz/media_attachments/files/116/968/602/918/356/214/original/a54ddeabbf2b0b49.jpg

2026-06-27 15:30:49 UTC

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Regarding the hospital iPhone failures caused by helium leaks mentioned in several comments，⬇
https://media.mathstodon.xyz/media_attachments/files/116/822/781/171/549/545/original/519b91f32a00f748.jpg

2026-06-02 18:19:12 UTC

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Sorry to go a bit off-topic, but I just came across a newly published paper by Japanese academics on the algebraic analysis of counterpoint. I thought you might find it interesting.https://arxiv.org/abs/2606.01102

2026-04-23 13:15:01 UTC

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Before clicking to read, I thought one of the motives was related to the incident a month ago when 200 containers of helium were stranded in the Persian Gulf, disrupting one-third of the global helium supply.

2025-09-15 04:59:17 UTC

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The day I read that headline, I briefly thought they had discovered something like a naked singularity😅

2025-08-15 07:07:27 UTC

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Apparently the American constitution is just too barbaric. The Putin–Medvedev commutativity law at least has a touch of mathematical elegance

2025-02-13 12:00:43 UTC

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Oh sorry，I found this information from a Chinese source（In the report about the early award-winning film 菊豆 'Ju Dou' by the famous Chinese director 张艺谋Zhang Yimou,it was briefly mentioned that the award is dedicated to Hugo Haas）. However, after seeing your reply, I tried verifying it with English sources, but couldn't find any. It's possible that the information reported by Chinese movie news writers might have been based on hearsay also.

2025-02-13 11:27:52 UTC

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Hugo Haas