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There's a news story going around (see for example [1]; other accounts are even more breathless) about an amateur mathematician, Yu Jianchun, finding an "alternative method to verify Carmichael numbers". I'm curious as to the method; does anyone know about this?

I don't think this is smoke and mirrors, as was the claimed proof of the Riemann Hypothesis by Opeyemi Enoch of the Federal University Oye-Ekiti, Nigeria. [2] Cai Tianxin of Zhejiang University has apparently read the proof and invited Yu Jianchun to give a talk at a graduate seminar, and CNN apparently ran the story by William Banks who apparently compared the discovery to the proof of the infinitude of Carmichael numbers (though from what appears in the article it seems he did not have access to the work itself).

But the article (appropriately, I suppose, for its nontechnical audience) doesn't give any real information on the discovery itself. The closest it comes is a picture of a letter (in Chinese) from Yu Jianchun with most of a formula: $$ \frac{\left(\frac{N^{P_1}-N}{N} - \frac{N^{P_1-P_2+1}-N}{P_1-P_2}\right)(P_1-P_2)}{P_2}(N,P_1,P_2 $$

(note that the rightmost part is cut off). There is also what might be a worked example, starting $$ \begin{align} =&\frac183^{3n}(3^n+1)^3(3^n+2)^3(3^{2n-1}+2\cdot3^{n-1}+1)^3\\ =&\left[\frac123^n(3^n+1)(3^n+2)(3^{2n-1}+2\cdot3^{n-1}+1)\right]^3 \end{align} $$ but it's hard to make sense of this without understanding the surrounding text (in Chinese).

[1] China's 'Good Will Hunting?' Migrant worker solves complex math problem

[2] Riemann Hypothesis not proved

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    $\begingroup$ a quote to put this in some perspective: Six professors and advanced students in Zhejiang University's math department listened to Yu's lecture. Some of the experts considered Yu's idea to be a "novelty", while some said "his results have a certain depth". Professor Cai decided to include Yu's formula with his latest work in English, and he gave Yu a book to help the logistics worker in further study. --- One might write to professor Cai for more info. $\endgroup$
    – Carlo Beenakker
    Commented Jul 21, 2016 at 13:47

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Apparently it is an alternative proof of the infinitude of Carmichael numbers.

The other proof mentioned in the articles ("done by academics 20 years ago") is:

As for the method, unless someone took notes of the lecture at Zhejiang University, I guess it is unpublished at this point (except for the fragments shown in the news).

As a side note, notice two equations shown in one the photos:

enter image description here

$$1. \quad(6n+1)(18n+1)(54n^2+12n+1)$$ $$2. \quad(1764n-139)(2268n+179)(1000188n^2-157752n+6221)$$

This looks very similar to Chernick's result that $(6n+1)(12n+1)(18n+1)$ is a Carmichael number if each of the factors is prime. It is open whether this family of Charmichael numbers is infinite or not.

This is pure speculation, but the non-linear factors in Yu's numbers might make a big difference, since that is not a Dickinson's conjecture-type problem anymore.

Added data for very small values ($n<11$) for the first expression. $C_i$ indicates the $i$-th Carmichael number.

\begin{array}{|c|c|c|c|} \hline \mathrm{n}& \mathrm{eq. 1 } & \\ \hline 1 & 7 \cdot 19\cdot 67 &8911=C_7\\ \hline 2 & 13 \cdot 37\cdot 241 &115921=C_{18}\\ \hline 3 & - &-\\ \hline 4 & - &-\\ \hline 5 & - &-\\ \hline 6 & 37 \cdot 109\cdot 2017& 8134561=C_{93}\\ \hline 7 & 43 \cdot 127\cdot 2731& 14913991=C_{125}\\ \hline 8 & - & -\\ \hline 9 & - & -\\ \hline 10 & 61 \cdot 181\cdot 5521& 60957361=C_{209}\\ \hline \end{array}

Edit. See the comment of Zhiyun Cheng below for a translation of the text in chinese.

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    $\begingroup$ Is it? Banks mentions the AGP result, but earlier in the article it's said to be a method to verify Carmichael numbers, and Yu refers to an "algorithm", which sounds more like "is n a Carmichael number or not" than "there are infinitely many Carmichael numbers". $\endgroup$
    – Charles
    Commented Jul 21, 2016 at 15:56
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    $\begingroup$ It's certainly a very difficult unsolved problem to show that there are infinitely many n for which each of the three factors in (1), or in (2), are prime - if that is what is required here. $\endgroup$
    – Ben Green
    Commented Jul 21, 2016 at 20:38
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    $\begingroup$ I can confirm that the number you ask about is not Carmichael. $\endgroup$
    – Gerry Myerson
    Commented Jul 22, 2016 at 0:56
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    $\begingroup$ Here is a translation of the picture above: PS: Absolute (Fermat) pseudoprimes that can be written as the product of three (prime?) factors can be completely classified. For example, the two simplest formulas which contain the quadratic term are 1. $(6n+1)(18n+1)(54n^2+12n+1)$; 2. $(1764n-139)(2268n-179)(1000188n^2-157752n+6221))$. 7 is a magic number. A deformation of the Fermat's little theorem (with the same base) $\frac{(\frac{N^{p_1}-N}{N}-\frac{N^{p_1-p_2+1}-N}{p_1-p_2})(p_1-p_2)}{p_2}(N, p_1, p_2$ $\endgroup$
    – Zhiyun Cheng
    Commented Jul 23, 2016 at 6:37
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    $\begingroup$ @Charles: $=\frac{1}{8}\times3^{3n}(3^n+1)^3(3^n+2)^3(3^{2n-1}+2\times3^{n-1}+1)^3$ $=[\frac{1}{2}\times 3^n(3^n+1)(3^n+2)(3^{2n-1}+2\times3^{n-1}+1)]^3$ The number is $y-x+1=(3^n+1)^3$ Therefore $\sum\limits_{i=0}^{(3^n+1)^3-1}(\frac{1}{2}\times3^{n-1}(3^{3n}+3^{2n}-5\times3^n-9)+i)^3=\frac{1}{2}\times3^n(3^n+1)(3^n+2)(3^{2n-1}+2\times3^{n-1}+1)$ $\endgroup$
    – Zhiyun Cheng
    Commented Aug 2, 2016 at 3:37
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For example, the two simplest formulas which contain the quadratic term are ...

I found simpler formulas: $$(6n+1)(12n+1)(24n^2+6n+1)$$ $$(1008n-79)(2268n-179)(326592n^2-51372n+2021)$$

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    $\begingroup$ Agree. A quote from a facebook friend of mine: "(to say that it is too short to publish in a paper) because its proof is homework-level. The result is that if $6k+1, 18k+1,54k^2+12k+1$ is prime, then their product $n$ is a Carmichael number. But one can prove this directly using Chinese remainder and little Fermat. Just need to show that $(n-1)$ is divisible by $6k, 18k, 54k^2+12k$." One can generate many similar criteria with a program. $\endgroup$
    – Ted Mao
    Commented Jul 27, 2016 at 2:04
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On page 136 of Cai Tianxin, The Book of Numbers, it says,

In 2015 the Chinese amateur, Jianchun Yu, who was then a packer in a bookstore warehouse in Hangzhou found that all the numbers like $(6k+1)(12k+1)(24k^2+6k+1)$ are Carmichael numbers if these three factors are all primes.

The proof is left as an exercise on page 142. I found no other mention of Jianchun's work in the book.

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  • $\begingroup$ Jianchuan works on higher order congruences by intuition and mental arithmetic, at least one mathematician trying to prove his formulas (from latest report 19-09-26 on Jianchuan, Jianchuan is a data analyst at Southeast Asia at present) $\endgroup$
    – Mike
    Commented Oct 6, 2019 at 14:06
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    $\begingroup$ @Mike what does "by intuition and mental arithmetic" mean? Is this a way of saying "not proving things"? $\endgroup$
    – Yemon Choi
    Commented Oct 6, 2019 at 18:37
  • $\begingroup$ @Yemon Choi Lack of details, but I don't think so. $\endgroup$
    – Mike
    Commented Oct 7, 2019 at 0:20
  • $\begingroup$ @Mike, what report? $\endgroup$
    – Gerry Myerson
    Commented Oct 7, 2019 at 1:13
  • $\begingroup$ @Gerry Myerson There' s many reports in Chinese, today I see a report say: Yitang Zhang say, I read his papers, I feel he is right. $\endgroup$
    – Mike
    Commented Oct 7, 2019 at 1:38
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Yu Jianchun's formulas: $$n=(6k+1)(18k+1)(54k^2+12k+1)=5832^4+2592k^3+450k^2+36k+1$$ $(n-1)/(6k)=972k^3+432k^2+75k+6$$ $$(n-1)/(18k)=324k^3+144k^2+25k+2$$ $$(n-1)/(54k^2+12k)=108k^2+24k+3$

Hope this helps.

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