Okay, I don't understand what the funny "O" thing is on the end of that.

Secondly, given this information, the assumption, and the proof, can we show the first 3 or 4 primes of this as described herein?

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- July 21st 2010, 10:46 AMSamson
- July 21st 2010, 10:52 AMundefined
The "O" is for ring of integers, it appears in the Wikipedia article

Quadratic integer - Wikipedia, the free encyclopedia

and I used it so I wouldn't have to specify omega.

As to your second question, when we choose some D, then we can find primes accordingly, but how about trying to do a little of this on your own? For example, if you want D congruent to 1 (mod 4), then you can choose D = 5 and try to find some primes, using the techniques that were used for Guassian and Eisenstein integers in the references I supplied, or if they don't apply, try to see why and try to develop a method.

Further notes to help you get started: I would start by just looking for irreducibles first and worrying about primes later. Also you should be able to see that a composite over the integers must be non-prime over your quadratic integer ring. Last, I would start by exhaustively searching using small pairs of (a,b). (Multiply them together, and you know that the product can't be irreducible, and if you can't produce an integer as such a product then you've found an irreducible.) - July 21st 2010, 11:06 AMSamson
I observed the links you suggested, but of them didn't help me in terms of finding primes that are applicable to my question (of having both a and b be either both even or odd with d>0). The examples that they gave were ones that you had repeated a bit in the first page of this thread, and also the page on Eisenstein primes yielded much confusion and not much direction on its integration with Quadratic Fields.

I really appreciate your help thus far, but this juncture is a little over my head given this documentation. If you can, I'd really appreciate it if you can continue your solution and explanation if you have sufficient time. - July 21st 2010, 11:12 AMundefined
Well, you can get started by just plugging in numbers.

Here I'll just make up a short hand, we'll say is represented by (a,b).(c,d) where and .

Fix D = 5.

Multiply

(1,1).(1,1)

(1,1).(1,2)

(1,1).(1,3)

(1,1).(2,1)

(1,1).(2,2)

(1,1).(2,3)

(1,1).(3,1)

(1,1).(3,2)

(1,1).(3,3)

(1,2).(1,1)

(1,2).(1,2)

(1,2).(1,3)

(1,2).(2,1)

(1,2).(2,2)

(1,2).(2,3)

(1,2).(3,1)

(1,2).(3,2)

(1,2).(3,3)

Etc.

Look for patterns, see what produces integers and what does not (if you know in advance you won't get an integer, don't bother multiplying), and generalize based on your results, and if there's a pattern that looks like it can be proven, attempt a proof.

EDIT: Well none of the products above will give integers because you need at least one negative sign somewhere, but the overall idea still stands. Also look at the general expansion of (a,b).(c,d) to see what choices could possibly result in integers. - July 21st 2010, 04:35 PMSamson
Well, I definitely tried. I tried so much that I missed your note about none of them giving you integers after I troubleshooted a billion times and never got one to come out! I came on here to say that they never came out right, and voila, you have your edit. Oh well...

Anyways, I've producted those two expressions together (what I call a general expansion) and I don't know really what I'm supposed to make of it that will lead me to real numbers (the first 3 or four primes where a,b are either both odd or even). Sorry that I'm lost here! - July 22nd 2010, 12:42 PMundefined
Yeah sorry about that mix-up. We can do

In order to have an integer, (ad + bc) must equal 0. There is not much point letting a = 0 because then we would end up with a multiple of 5 which we already know is composite. Likewise there's no use letting b = 0 because then we have either a = 0 or d = 0, and if d = 0 then we are just multiplying two integers.

While I'm sure we can analyse this further, I took the easy way out (in my opinion) and wrote a brute force program. This is in Python, should run on any version

Code:`import math`

def isprime(n):

if n < 2:

return False

# just trial division

if n%2 == 0:

return False

for i in range(3, int(math.floor(math.sqrt(n))+1), 2):

if n%i == 0:

return False

return True

for a in range(1,100):

for b in range(-75,75):

if b != 0:

for c in range(-75,75):

if (b*c) % a == 0:

d = -1*b*c / a

n = a*c+5*b*d

if n > 0 and isprime(n):

print "!"+str(n),a,b,c,d

A snippet of the output:

Code:`!89 6 -5 -6 -5`

!3 6 -3 -2 -1

!31 6 -1 6 1

!31 6 1 6 -1

!3 6 3 -2 1

!89 6 5 -6 5

So it looks like the first irreducible is 2, and the next irreducible isn't until 53! - July 23rd 2010, 10:06 AMSamson
- July 27th 2010, 03:16 AMSamson
- August 4th 2010, 07:35 AMundefined
- August 5th 2010, 06:05 AMSamson
- August 9th 2010, 04:10 AMSamson
Did you by chance get that reformulation done? Once again, I really appreciate the help!

- September 4th 2010, 01:48 PMundefined
Long time delay, busy month, sorry for that. I think this is the last thread of yours that I was active on that has a loose end.

Mostly just read my comments in this post that I made just now.

I notice that when I add the condition a%2 == b%2 (I also tried abs(a%2) == abs(b%2)) to my Python program it produces no output whatsoever. Perhaps this means that the primes of are never irreducible in this ring. However I would have to look more deeply and don't have too much time. I recommend looking at the PDF on arXiv referenced in the post linked to above, since you wanted primes instead of irreducibles anyway. - September 5th 2010, 10:47 AMSamson
- September 5th 2010, 11:58 AMundefined
Actually I think I interpreted lack of output oppositely from what I should have; assuming no bugs (which is an assumption that should definitely be checked), no output means that

**all**the primes of Z are irreducible in the ring, because it means there is no way to multiply two algebraic integers to get a prime in Z. It had been so long since I wrote the program, I forgot what it did..

Edit: Aha, the program is buggy in the sense that it had a false assumption leading to it; I wrote algebraic integers as rather than . I now remember noticing this earlier but by the time I made the response had forgotten. Anyway I take responsibility for the mistakes I make along the way but can't help but feel like I'm doing 99% the work here. Is it really much of an enlightenment for you to be shown the answers for these problems? Won't you get more out of it by finding the answers*yourself*? - September 5th 2010, 03:23 PMSamson
I definitely have been trying to find them. I've been working over a month on some of these, but I'm coming here so someone can show me how its done so I can learn from it. I'm learning out of a book, and I can't ask the book questions or ask it to solve my examples. Essentially having someone to show me how these work is crucial in helping me understand the rest of the example problems (which I always work on alone).

So how does modifying your equation by the factor of 1/2 change things?