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.)
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.
Here I'll just make up a short hand, we'll say is represented by (a,b).(c,d) where and .
Fix D = 5.
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.
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!
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
The code isn't written to be elegant, just to require as little thought as possible to code.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:
This means for example we can get product 3 by letting (a,b,c,d) = (6,-3,-2,-1). I added the exclamation marks just to make it easier to search.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!
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.
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?
So how does modifying your equation by the factor of 1/2 change things?