20050912, 02:17  #1 
Cranksta Rap Ayatollah
Jul 2003
641 Posts 
Prime generating polynomials that stop?
Are there any polynomials that generate a finite number of primes?

20050912, 03:17  #2 
Dec 2003
Hopefully Near M48
1758_{10} Posts 
p(x) = 2
Maybe you would like to clarify your question some more? 
20050912, 07:51  #3 
"Nancy"
Aug 2002
Alexandria
2,467 Posts 
Challenging me to the title of Captain Obvious?
Alex 
20050912, 17:44  #4 
Cranksta Rap Ayatollah
Jul 2003
641 Posts 
Okay, egg on my face ;)
I was thinking about n^{2}+n+41, and as far as I know, it is unknown whether it produces an infinite number of primes. I was thinking that maybe there would be some value excluding trivial solutions, but I'm not entirely sure what would constitute "trivial" and think I just asked a bonehead question 
20050912, 18:16  #5 
Sep 2004
5×37 Posts 
Hi !
Just a related question (but the converse of TravisT absence of primes): the Ulam spiral produces diagonals rich in primes corresponding to 2nd polynomials : an^2+bn+c...In a book called "merveilleux nombres premiers" by JP Delahaye, it is written that this isn't well understood...can you give me some links or explanations about it? Thanks. Philippe. Last fiddled with by Phil MjX on 20050912 at 18:17 
20050912, 19:47  #6  
Bamboozled!
"๐บ๐๐ท๐ท๐ญ"
May 2003
Down not across
10110000100110_{2} Posts 
Quote:
Give "trivial" a nontrivial definition and the question is probably not boneheaded. Here is a quadratic that generates only one prime number when evaluated at integer arguments: p(x) = 2x^2. Note that, unlike the earlier example, this one generates an infinite number of different values but only one is prime. Paul 

20050912, 20:01  #7 
Jun 2005
Near Beetlegeuse
2^{2}×97 Posts 
Coincidentally, there is an interesting connection between polynomials that produce primes and Ulam spirals. In Paul Hoffmanns book The Man Who Loved Only Numbers a biography of Paul Erdos there is an explanation of how Stan Ulam came to discover these spirals. He was doodling while listening to a long boring paper at a math conference, and wrote the number 1 in the middle then wrote successive integers in a counterclockwise square spiral.
17 16 15 14 13 1854312 1961211 2078910 and so on. He noticed that the primes fall on common diagonals. After playing with this idea he found that if you start in the middle with 17 instead of 1 and do the same thing, you end up with a long diagonal containing the primes 227, 173, 127, 89, 59, 37, 23, 17, 19, 29, 47, 73, 107, 149, 199, 257. And the connection with polynomials? Well, the polynomial n^2 + n + 17, discovered by Euler, produces these exact same primes. Curious, but true. I should add that this only works for n = 0 to 15 Last fiddled with by Numbers on 20050912 at 20:06 Reason: Added range for n 
20050912, 22:49  #8  
Cranksta Rap Ayatollah
Jul 2003
641 Posts 
Quote:


20050912, 23:38  #9  
Nov 2003
2^{2}×5×373 Posts 
Quote:
infinitely many examples of this type. f(x) = (x1) (x6) or f(x) = (x3)(x2)(x5)(x7) etc. 

20050913, 00:26  #10 
Jan 2005
Caught in a sieve
5×79 Posts 
I'm assuming you're looking for f(x), where for x any nonnegative integer, there are only N>1 cases where f(x) is prime?
Okay, try this on for size : f(x) = 3x f(0) = 3, Prime! f(1) = 2, Prime! f(2) = 1, not prime! f(3) = 0, not prime! for x>=4, f(x) < 0; therefore f(x) is not prime! Last fiddled with by Ken_g6 on 20050913 at 00:28 Reason: "whole number" is not well defined. 
20050913, 00:51  #11 
Aug 2003
Upstate NY, USA
2·163 Posts 
f(x) = n^2+n+41  X*floor( (n^2+n+41)/X )
where X is some bound that we do not want our function to produce values above probably not exactly what you're looking for though... 
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