Timeline for A Diophantine equation involving prime powers.
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Apr 2, 2013 at 16:18 | comment | added | user9072 | In particular, the presumably smallest mentioned above is not from the original paper but was found by Selfridge. | |
| Apr 2, 2013 at 16:15 | comment | added | user9072 | @Noam D. Elkies: this seems close to the original proof; the paper is a page.and a half. Specifically, what is shown is that each a congruent 1 mod $(2^{32}-1)641$ and -1 mod the other (than 641) proper divisor of F_5 is an admissible choice. And, this is done via always finding a divior from the list you give. | |
| Apr 2, 2013 at 15:33 | comment | added | Noam D. Elkies | I remember this from 30+ years ago when one of the USAMO problems asked in effect to prove that a Sierpiński number exists. The usual solution (among the few who solved it at all) was to use covering congruences to find $a$ such that $2^n a + 1$ is always a multiple of one of $3$, $5$, $17$, $257$, $65537$, and the two factors of $2^{32}+1$ (so this ties in with the OP's first problem too...). | |
| Apr 2, 2013 at 14:17 | comment | added | user9072 | And to avoid a misconception I should perhaps stress that the 'one can show' only means that it was possible to do this rather recently (a bit more than ten years ago) for leading experts (Friedlander--Iwaniec and Heath-Brown, resp.) | |
| Apr 2, 2013 at 13:53 | comment | added | user9072 | By contrast the number of the form you ask for are about log N of all numbers up to N. This is way too sparse for todays tools. This is for what one can prove. For what one expects the heuristic is: exclude all members of your set that are not prime for some 'obvious' reason (such as congruence conditions) and then sum the reciprocals of the logarithms of the numbers in your set. If this diverges you expect inifnitely many primes, if it converges you expect at most finitely many. [While I say 'obvious' the reason might not be that obvious in some cases of course.] | |
| Apr 2, 2013 at 13:47 | comment | added | user9072 | You are welcome! Yes, such things are essentially always open. A rough meta-heuristic: if you define a set in some way [that does not somewhat directly imply that it contain infinitely many prime numbers (or powers, this does not change that much)] than you only have a chance to show it does contain infinitely many prime numbers (or powers) if the set is somewhat dense. For example, one can show that there are infinitely many primes of the form x^2 + y^4 or also x^3+2y^3 but the set of all such numbers up to N is about N^(3/4) and N^(2/3). This is about the limit of current technology. | |
| Apr 2, 2013 at 13:15 | comment | added | Uep | Thank you, quid! That is very interesting. Now I wonder if there is a fixed number $a$ and a fixed prime $p$ such that the set $\{{ap^n+1\}}$ consists of infinitely many prime powers. Based on what you have said, I feel that this question is also open. | |
| Apr 2, 2013 at 13:11 | vote | accept | Uep | ||
| Apr 2, 2013 at 11:50 | comment | added | user9072 | It might also be worth noting that there are even prime Sierpiński numbers, such as 271129. Also the year is (of course) 1960 not 1060. | |
| Apr 2, 2013 at 11:42 | comment | added | user9072 | I do not want to edit for this (right away), but the 'just like for Mersenne primes' is not quite precise, as in the the Mersenne case one also has a restriction for the $n$ (it has to be prime) from a general factorization. Yet the restriction is sufficiently weak that the series over the reciprocals of the 'admissible' exponents still diverges. Moreover, some restriction (excluding certain residue classes for teh expoenent) will typically be present; yet again it will be/can be only weak so that the series over the 'admissible' ones stays divergent. | |
| Apr 2, 2013 at 11:27 | history | answered | user9072 | CC BY-SA 3.0 |