 mersenneforum.org prime producing polynomial 2
 Register FAQ Search Today's Posts Mark Forums Read  2016-03-23, 17:06   #23
chris2be8

Sep 2009

32·227 Posts Quote:
 Originally Posted by CRGreathouse This is a good question. Let f(x) be the putative prime-producing polynomial. gcd(x,1) = 1 for all x, so f(1) = p is prime. But then f(p + 1), f(2p + 1), ..., are all divisible by p, since each power of x will be 1 mod p. So they can be prime only if f(1) = f(p + 1) = f(2p + 1) = ..., which is possible only if f is the constant polynomial f(x) = p.
Thanks for that. I had said there could be a few other types of exception. But generalizing your argument:

Pick any n such that f(n)=p where p is prime. Then f(n+p), f(n+2p), etc will all be divisible by p since each power of x will be the same mod p as it is for f(n). So most values of n will generate a composite (unless it's a constant polynomial which can only generate 1 prime).

Chris   2016-03-24, 19:01   #24
ET_
Banned

"Luigi"
Aug 2002
Team Italia

113168 Posts Quote:
 Originally Posted by paulunderwood The quadratic x^2+1 has not been proven to generate an infinite number of primes. See 30 minutes into: https://www.youtube.com/watch?v=rwH-5VhBPGc and if you have the time the whole video is worth watching. Oh I see x is not an integer???
Wonderful video, thaank you    2016-03-26, 01:16   #25
MattcAnderson

"Matthew Anderson"
Dec 2010
Oregon, USA

24×32×5 Posts Hi Math People

Attached are slides for my write-up for n^2 + n + 41. I have removed all reference to the word bifurcation. That was incorrect. The points in the graph of divisors lie on parabolas.

Regards
Matt
Attached Files A prime producing quadratic expression 4.pdf (428.4 KB, 212 views)   2016-04-19, 08:35 #26 bhelmes   Mar 2016 52·13 Posts A peaceful day for all members there are some "new" results for the polynomial f(n)=n^2+n+41 for n<=2^35 http://www.devalco.de/basic_polynomi...a=1&b=1&c=41#7 For people who are not familiar with quadratic prime generators: It is possible to make a sieving construction for quadratic irreducible polynomials like f(n)=n² +1. The Sieve of Eratosthenes is a more specific variation of this construction. Normally the primes with p=f(n) or p|f(n) appear double periodically on the polynomial. If you divide the appearing primes you can be sure that there only rest a prime or the number one. Would nice to have a little feedback. The topic is quite interesting for people who look for some prime generators. An overview is in the web under http://devalco.de/#106 Have a lot of fun with the primes Bernhard   2016-04-30, 07:47 #27 MattcAnderson   "Matthew Anderson" Dec 2010 Oregon, USA 13208 Posts Hi Math People, Thank you for considering my little project. As I have state before the word "bifurcation" was incorrectly used by me to describe a graph. New words are graph of discrete divisors. For what it is worth. Also, Bernhard, your work on seiving and quadratic functions is very interesting. Regards, Matthew Attached Thumbnails   2017-04-25, 12:42   #28
MattcAnderson

"Matthew Anderson"
Dec 2010
Oregon, USA

10110100002 Posts primes of the form n^2+n+41

Hi all,

I continue to doodle with h(n). I define h(n) as n^2 + n + 41. We assume n is a positive integer. So far we know that no primes less than 40 ever divide h(n). Also, there are no positive integers n that make 59 divide h(n).

We prove these two facts by exhaustive search in a residue table.

Some of my progress is shown at this webpage -

We know that if n is congruent to x mod y and (x,y) is on the curve

p(r,c) = (c*x – r*y)2 – r*(c*x – r*y) – x + 41*r^2 = 0

and 0<r<c, c>1 and gcd(r,c) = 1 and all four of r,c,x, and y are integers,

then h(n) is a composite number.

further, all n such that h(n) is a composite number are probably on p(r,c)=0.

Each such pair (r,c) yields integer points on a parabola.

Possible next steps -
We want to show that h(n) is prime an infinite number of times. p(r,c) is 0 for an infinite number of values x and y. Given that x and y are counting numbers and restrict x and y to be not a solution of p(r,c) = 0, can we infer that there are still an infinite set of pairs (x,y) such that n = x mod y will give us a prime value for h(n)?

This seems to be a hard question.

Regards,
Matt
Attached Files extended resedue table.pdf (118.3 KB, 175 views) extended resedue table explaination.txt (415 Bytes, 154 views) prime producing polynomial continues.pdf (148.1 KB, 170 views)   2017-10-16, 00:47 #29 MattcAnderson   "Matthew Anderson" Dec 2010 Oregon, USA 24×32×5 Posts Hi all, There are two webpages that I have made regarding Prime Producing Polynomial. The webpages are - sites.google.com/site/mattc1anderson/prime-producing-polynomial sites.google.com/site/primeproducingpolynomial Regards, Matt   2020-11-03, 22:58 #30 MattcAnderson   "Matthew Anderson" Dec 2010 Oregon, USA 24·32·5 Posts Hi again all, I added a counter example to a conjecture of mine that i about 2 years old. Regarding our Prime Producing Polynomial Project for the enhancement of mathematical trivia database Specifically, counterexample to conjecture about x2_plus_x_plus_41.pdf Cheers, Matt Last fiddled with by MattcAnderson on 2020-11-03 at 22:59   2021-01-31, 19:03 #31 MattcAnderson   "Matthew Anderson" Dec 2010 Oregon, USA 24×32×5 Posts Hi all, Today I coded a Maple worksheet and put it on the internet. The file is called "an interesting graph with negatibes done.pdf". You can click here to see. Regards, Matt   Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post carpetpool Miscellaneous Math 14 2017-02-18 19:46 MattcAnderson MattcAnderson 4 2016-09-06 14:15 MattcAnderson MattcAnderson 28 2014-04-26 01:35 MattcAnderson MattcAnderson 0 2013-07-19 04:25 R.D. Silverman NFSNET Discussion 13 2005-09-16 20:07

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