Prime generating polynomials that stop? - Page 3

Numbers · 2005-09-14, 19:01

Quote:

Originally Posted by R.D. Silverman

Typical newbie crank response: Open mouth and display ignorance.

-5 is prime. It differs from +5 only by multiplication by a UNIT.

You might understand this if you ever bothered to STUDY this subject.

Mr Silverman,

I did not deserve this kind of response.

http://www.mersenneforum.org/showthread.php?t=4656
http://www.mersenneforum.org/showthread.php?t=4508
http://www.mersenneforum.org/showthread.php?t=4620
http://www.mersenneforum.org/showthread.php?t=4367
http://www.mersenneforum.org/showthread.php?t=4364

I think the evidence will show that I am �bothering� to study this subject. But the whole subject can not be absorbed overnight. Learning maths is not the work of a few weeks or months, but of a lifetime. Each new discovery made along the way opens up new vistas for us to explore and makes us realise how much more there is to learn, to know, to understand.

I love maths. I just love that feeling I get when the light bulb comes on and the solution to something that has puzzled me for weeks is suddenly as obvious as the clouds in the sky. I know that I will never prove or discover anything new or publish a paper. But I love it, even though at times I feel that it does not love me.

Would you please be so kind as to post at least one reference to a knowledgeable source that has a definition of prime number that does not include the words �positive integer.�?

R.D. Silverman · 2005-09-14, 19:54

Quote:

Originally Posted by Numbers

Mr Silverman,

I did not deserve this kind of response.

Yes. You did. A knowledgeable person (Xilman) told you that -5 was prime.

Rather than accept this or ask why he made the statement, you had to
argue back. That makes you a crank.

Pick up any book on algebraic number theory. Or any book on ring theory.
There are (I am simplifying somewhat) 3 types of numbers in a ring:
primes, composites and units. A unit is an element that has a multiplicative
inverse in the ring. Within Z, +1 and -1 are units. Primes whose norms are the same, but differ by multiplication by a unit, are called associates. But
both are prime.

Under your imagined definition, e.g. the integer -10 can not be factored at all
into primes!

Look up the words "prime ideal". The "prime is a positive integer" definition
will only be found in elementary number theory books. Once one gets away
from Z, the very meaning of "positive" vs. "negative" is lost.

Consider, for example, the Gaussian integers. 1-i is prime in this ring.
But it is neither positive or negative. -1 + i is an associate. It too is prime.

Numbers · 2005-09-14, 21:03

Quote:

Originally Posted by R.D. Silverman

Rather than accept this or ask why he made the statement, you had to argue back. That makes you a crank.

Well I�m sorry Mr Silverman but that is just not true. The first sentence of my post ends with a question mark, which by any normal standards means it is a question. IOW I am asking why, just as you suggested I should.

Having asked what Xilman meant by �reasonable� definition I supplied one that is not a million miles from the one found on Mathworld. All I then said was that by this �reasonable� definition it would appear that �5 is not prime. This is, in the language of rhetoric that I would have thought an intelligent person such as yourself or Xilman would easily understand, an invitation to answer the previously explicitly stated question �What do you mean by reasonable definition�?

If I had said, �xilman you loser, everyone knows negatives aren�t prime�, then I would have deserved any amount of opprobrium. But having asked the question you suggested I should, how do you now interpret that as arguing back?

xilman · 2005-09-15, 09:02

Quote:

Originally Posted by Numbers

Well I�m sorry Mr Silverman but that is just not true. The first sentence of my post ends with a question mark, which by any normal standards means it is a question. IOW I am asking why, just as you suggested I should.

Having asked what Xilman meant by �reasonable� definition I supplied one that is not a million miles from the one found on Mathworld. All I then said was that by this �reasonable� definition it would appear that �5 is not prime. This is, in the language of rhetoric that I would have thought an intelligent person such as yourself or Xilman would easily understand, an invitation to answer the previously explicitly stated question �What do you mean by reasonable definition�?

If I had said, �xilman you loser, everyone knows negatives aren�t prime�, then I would have deserved any amount of opprobrium. But having asked the question you suggested I should, how do you now interpret that as arguing back?

I'm on Numbers side in this particular argument. He did indeed ask for my definition of "reasonable" and pointed out how his criterion for reasonableness had some conflict with my earlier statement.

Unfortunately, I was unable to provide a definition quickly enough before tempers became inflamed.

Bob has since provided a defintion in words close to what I would have used. I, however, would have added that zero is not prime, composite or a unit.

Paul

Numbers · 2005-09-15, 09:17

I�ve been thinking about this, probably more than is healthy for me, but� and I have come up with what I think is a perfectly plausible scenario.

You were skimming down the thread and saw the words �-5 is not prime� and without reading the rest of the post you went into newbie-crank-overdrive mode and flashed off your critical rejoinder.

Then, when I suggested that I had not deserved such a verdict, you were sufficiently confident of your own judgement that you did not go back and actually read the post, you simply confirmed your previous assessment. IOW you probably still haven�t read my post properly, which means that you made a simple mistake. The sort of thing any one of us could do any day of the week.

Assuming this is true, or at least pretty close, it would be far too much for me to expect that you would admit it in the forum. So I am just going to ask you to admit to yourself that you made a perfectly simple mistake. Then, hopefully, next time someone else makes a simple mistake you will perhaps be just a little less hasty to be so withering with your contempt.

Now let�s move on and talk some maths.
Apart from the two equations mentioned already, in 1772 Euler discovered a little family of equations that produce a finite number of primes. They all take the form n^2 + n + q where q = {2, 3, 5, 11, 17, 41} and produce primes for n > -1, n < q-1.

Regarding jinydu�s question for more information about the tests run on Maniac II, I have no more than I posted before, sorry.

Numbers · 2005-09-21, 12:45

If you want to see how these Ulam spirals are generated I found this:
http://www.alpertron.com.ar/ULAM.HTM

It's on Dario Alpertron's site and has a cool applet that generates Ulam spirals and shows the polynomials for the diagonals and which primes are associated with them. You can also zoom in and out, change the start number and do all sorts of other weird stuff.

Good work Dario,

jinydu · 2005-09-23, 01:52

If $n^2+n+41$ really does generate infinitely many primes, this raises another question:

How "often" does it generate primes (asymptotically)?

In other words, if p(N) is the number of times $n^2+n+41$ is prime, for $1\leq{}n\leq{}N$ , what is

$\Large \lim_{N\to\infty} \frac{p(N)}{N}$

?

dave_dm · 2005-09-23, 17:31

Quote:

Originally Posted by jinydu

If $n^2+n+41$ really does generate infinitely many primes, this raises another question:

How "often" does it generate primes (asymptotically)?

In other words, if p(N) is the number of times $n^2+n+41$ is prime, for $1\leq{}n\leq{}N$ , what is

$\Large \lim_{N\to\infty} \frac{p(N)}{N}$

?

It's hard to determine the asymptotic proportion of primes generated by any polynomial, ignoring trivial cases. The f(n) = n case is bad enough (the answer being the Prime Number Theorem).

The reason that f(n) = n^2 + n + 41 generates primes for -40 <= n < 40 is that the number field $\mathcal Q(\sqrt{-163})$ has class number 1 (the 41 comes from (163+1)/4). The proof requires quite a bit of algebraic number theory to understand though. This result extends to all imaginary quadratic fields having class number 1, see http://mathworld.wolfram.com/GausssC...erProblem.html for a list. So for instance, f(n) = n^2 + n + 17 is prime for -16 <= n < 16.

Dave

R.D. Silverman · 2005-09-23, 18:11

Quote:

Originally Posted by dave_dm

It's hard to determine the asymptotic proportion of primes generated by any polynomial, ignoring trivial cases. The f(n) = n case is bad enough (the answer being the Prime Number Theorem).

The reason that f(n) = n^2 + n + 41 generates primes for -40 <= n < 40 is that the number field $\mathcal Q(\sqrt{-163})$ has class number 1 (the 41 comes from (163+1)/4). The proof requires quite a bit of algebraic number theory to understand though. This result extends to all imaginary quadratic fields having class number 1, see http://mathworld.wolfram.com/GausssC...erProblem.html for a list. So for instance, f(n) = n^2 + n + 17 is prime for -16 <= n < 16.

Dave

Actually, the density is given by the Bateman-Horn conjecture. (but
a proof is lacking)

#{p < x | p = n^2+n+41 } ~ C sqrt(x)/log(x) where the constant C
is given by this strange looking infinite product involving the Jacobi symbol
with respect to the discriminant (in this case -163)

Primeinator · 2005-09-23, 21:07

X^2 + X + 17 also generates a finite series of primes.

mfgoode · 2006-10-06, 17:16

Just got my 6th edition 2006 (C2007) of the book by D.M. Burton very cheaply as it is subsidised. Its titled 'Elementary Number Theory' and has been updated.
Apart from the controversy raging above, I give particulars, as I find that many posters are very misinformed.

First of all let me mention, and I have done so before in this forum, that it has been shown that no polynomial of the from n^2 + n + q with q a prime can do better than the Euler polynomial in giving primes for *succesive* values of n.

The current record holder, post 1988, produces a better string of 45 prime values than the ones considered.
Please check it out

k(n) = 36n^2 -810n + 2753.

It is 'PROVED' that there is no nonconstant polynomial f(n) with integral coefficients that takes on just prime values.

If anyone requires the proof ( a short one) I will give it on demand.

Mally

2005-09-15, 09:17	#27
Numbers Jun 2005 Near Beetlegeuse 2²·97 Posts	I�ve been thinking about this, probably more than is healthy for me, but� and I have come up with what I think is a perfectly plausible scenario. You were skimming down the thread and saw the words �-5 is not prime� and without reading the rest of the post you went into newbie-crank-overdrive mode and flashed off your critical rejoinder. Then, when I suggested that I had not deserved such a verdict, you were sufficiently confident of your own judgement that you did not go back and actually read the post, you simply confirmed your previous assessment. IOW you probably still haven�t read my post properly, which means that you made a simple mistake. The sort of thing any one of us could do any day of the week. Assuming this is true, or at least pretty close, it would be far too much for me to expect that you would admit it in the forum. So I am just going to ask you to admit to yourself that you made a perfectly simple mistake. Then, hopefully, next time someone else makes a simple mistake you will perhaps be just a little less hasty to be so withering with your contempt. Now let�s move on and talk some maths. Apart from the two equations mentioned already, in 1772 Euler discovered a little family of equations that produce a finite number of primes. They all take the form n^2 + n + q where q = {2, 3, 5, 11, 17, 41} and produce primes for n > -1, n < q-1. Regarding jinydu�s question for more information about the tests run on Maniac II, I have no more than I posted before, sorry. Last fiddled with by Numbers on 2005-09-15 at 09:25 Reason: Spelling

2005-09-23, 01:52	#29
jinydu Dec 2003 Hopefully Near M48 2·3·293 Posts	If $n^2+n+41$ really does generate infinitely many primes, this raises another question: How "often" does it generate primes (asymptotically)? In other words, if p(N) is the number of times $n^2+n+41$ is prime, for $1\leq{}n\leq{}N$ , what is $\Large \lim_{N\to\infty} \frac{p(N)}{N}$ ? Last fiddled with by jinydu on 2005-09-23 at 01:53

2006-10-06, 17:16	#33
mfgoode Bronze Medalist Jan 2004 Mumbai,India 804₁₆ Posts	polynomials and primes. Just got my 6th edition 2006 (C2007) of the book by D.M. Burton very cheaply as it is subsidised. Its titled 'Elementary Number Theory' and has been updated. Apart from the controversy raging above, I give particulars, as I find that many posters are very misinformed. First of all let me mention, and I have done so before in this forum, that it has been shown that no polynomial of the from n^2 + n + q with q a prime can do better than the Euler polynomial in giving primes for succesive values of n. The current record holder, post 1988, produces a better string of 45 prime values than the ones considered. Please check it out k(n) = 36n^2 -810n + 2753. It is 'PROVED' that there is no nonconstant polynomial f(n) with integral coefficients that takes on just prime values. If anyone requires the proof ( a short one) I will give it on demand. Mally

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2005-09-21, 12:45	#28
Numbers Jun 2005 Near Beetlegeuse 2²×97 Posts	If you want to see how these Ulam spirals are generated I found this: http://www.alpertron.com.ar/ULAM.HTM It's on Dario Alpertron's site and has a cool applet that generates Ulam spirals and shows the polynomials for the diagonals and which primes are associated with them. You can also zoom in and out, change the start number and do all sorts of other weird stuff. Good work Dario,

2005-09-23, 21:07	#32
Primeinator "Kyle" Feb 2005 Somewhere near M52.. 3·5·61 Posts	X^2 + X + 17 also generates a finite series of primes.