Timeline for Constructing prime numbers
Current License: CC BY-SA 3.0
19 events
| when toggle format | what | by | license | comment | |
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| Feb 4, 2023 at 12:25 | comment | added | Ali Taghavi | +1 for your interesting question. I am a beginer in number theory but I think the world of prime numbers is a mysterious world | |
| Sep 16, 2015 at 13:22 | history | edited | Denis Serre | CC BY-SA 3.0 |
added 13 characters in body
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| Dec 14, 2010 at 21:39 | answer | added | Asterios Gkantzounis | timeline score: 4 | |
| Nov 12, 2010 at 7:05 | comment | added | Denis Serre | @Michael. Good point. | |
| Nov 12, 2010 at 3:44 | comment | added | Michael Hardy | "The classic proof" is a modern proof. Euclid didn't consider the number that is 1 more than the product of the first $k$ primes. He considered the number that is 1 more than the product of an arbitrary finite set of primes, with no assumption that they were the smallest ones. | |
| Oct 13, 2010 at 15:28 | comment | added | Fedor Petrov | we may construct $p_{n+1}$ as minimal prime divisor $(\prod_{i=1}^n p_i)^k-1$ for $k=1,2,\dots$ :) | |
| Oct 13, 2010 at 13:26 | answer | added | Greg Kuperberg | timeline score: 8 | |
| Sep 16, 2010 at 6:41 | comment | added | Denis Serre | With $M$s and $N$s, one has $q_9=p_9=23$, because of $$2\cdot3\cdot5\cdot7\cdot11\cdot17\cdot19+13\equiv0\qquad(mod\,23).$$ However, Moret-Bailly's obstruction arises at the next step, because the product $Q=2\cdot3\cdots 23\equiv17$ (mod $29$), and none of $\pm17$ is a square (mod $29$).This proves that $q_{10}>p_{10}=29$. | |
| Sep 16, 2010 at 3:14 | answer | added | Charles | timeline score: 5 | |
| Sep 16, 2010 at 2:50 | history | edited | Charles |
edited tags
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| Sep 15, 2010 at 17:02 | comment | added | Denis Serre | Gasp ! I made another mistake. Replace $8m\pm1$ above by $4m+1$. The latter is the criterion for $-1$ to be a square (mod $p$), whereas the former is that for $2$ to be a square. | |
| Sep 15, 2010 at 16:10 | comment | added | Denis Serre | My penultimate comment is logically incorrect. What we should say is the following. Let us take in account the $M$s and the $N$s. In order that the obstruction raised par Laurent M.-B. works, it is necessary that $p_n$ be of the form $8m\pm1$. However, it is still possible that the first instance where $q_n\ne p_n$ is not caused by L. M.-B.'s obstruction. We only know that it is not larger than the first instance for which the obstruction arises. | |
| Sep 15, 2010 at 15:51 | comment | added | Denis Serre | With $M$s and $N$s, we do have $q_8=p_8(=19)$, because of $$2\cdot3\cdot11\cdot17+5\cdot7\cdot13=1577=83\cdot19.$$ | |
| Sep 15, 2010 at 15:42 | comment | added | Denis Serre | Sorry for the LaTeX fault. So, let us use both the $M$s and the $N$s. Let $n$ be maximal for the property that $q_k=p_k$ for all $k<n$. Then neither $\prod_{1≤j≤n−1}p_j$, nor its opposite $-\prod_{1≤j≤n−1}p_j$ can be a square (mod $p_n$).In particular, $-1$ is a square, and therefore $p_n$ is of the form $8m\pm1$.This leaves some hope that $q_8$ be equal to $19$ | |
| Sep 15, 2010 at 15:26 | comment | added | Denis Serre | Nice argument. Merci Laurent ! | |
| Sep 15, 2010 at 12:00 | comment | added | Laurent Moret-Bailly | With the Ms only, if $q_k=p_k$ for all $k\leq n$, then for some $I\subset[1,n-1]$ we have $\prod_{i\in I}p_i\equiv\prod_{i\not\in I}p_i\pmod{p_n}$. In particular $\prod_{1\leq i\leq n-1}p_i$ is a square mod $p_n$. This is true for $n<8$ but not for $n=8$ ($p_n=19$). Hence $q_8\neq p_8$. | |
| Sep 15, 2010 at 9:35 | comment | added | Denis Serre | Right. If you take in accound the $M$s and the $N$s, you may start from $k=1$. | |
| Sep 15, 2010 at 9:33 | comment | added | Laurent Moret-Bailly | You should start with $k=2$ and $q_1=2$, $q_2=3$. With your construction, $q_2$ is not defined. | |
| Sep 15, 2010 at 8:42 | history | asked | Denis Serre | CC BY-SA 2.5 |