20051224, 04:41  #1 
Nov 2005
182_{10} Posts 
What way would you find numbers with a set number of factors?
If prime numbers are numbers that only factor into 1 and themselves, then what would 3 factors including one and itself be?
It turns out that the middle factor must follow a rule: 1*n^2=n*n=x The factors of n^2 must be 1, n, n^2. Since n has to be prime, that means that the answer is the set of all the squares of primes. Try this for 4, 5, 6, ... factors for x including 1 and x. Last fiddled with by nibble4bits on 20051224 at 04:42 
20051229, 23:43  #2 
Aug 2003
Upstate NY, USA
2×163 Posts 
Let Pi be distinct primes for all i:
4 factors: (P1)^3 or (P1)*(P2) 5 factors: (P1)^4 6 factors: (P1)^5 or (P1)^2*(P2) 7 factors: (P1)^6 8 factors: (P1)^7 or (P1)^3*(P2) or (P1)*(P2)*(P3) so on, so forth... 
20051230, 21:01  #3 
Nov 2005
2×7×13 Posts 
At 1 and 0 you get the identies 1 and 0. 0 can't be divided at all and 1 has only one possible factor including itself and itself. Hehe we'll just say "period" to make more sense.
At 2 total factors there's only primes. (works both ways: A>B and B>A) The solutions for 2 factors are a kind of 'key' to the higherfactorcount sets in the 3D tree since obviously primes are the simplest factors possible. If there's an infinite number of primes, is there an infinite number of the 3factor results? 4factor? All? See 2nd post to see why there must be. Yes, there are infinite primes but no telling how long you'll have to wait to find the next one! I left this in the spoiler so those who want to can do the work themselves to find and understand the proof. 
20051230, 21:45  #4 
Jun 2005
Near Beetlegeuse
388_{10} Posts 
This is very closely related to something I was thinking about this afternoon.
As there are an infinite number of primes, and as all primes are either 1(mod 6) or 5(mod 6), are there an infinite number of primes 1(mod 6)? 
20051230, 21:53  #5 
"Nancy"
Aug 2002
Alexandria
2,467 Posts 
Yes, this is a special case of the "prime number theorem for arithmetic progressions." Simply put, it says that if you have an arithmetic progression a+b*x with gcd(a,b)=1 and x∈N, you get infinitely many primes. What is more, each such progression for different values of a (but the same b) gets an "equal share" of the primes. See Crandall and Pomerance, Prime Numbers, Theorem 1.1.5.
Alex 
20051230, 22:20  #6 
Jun 2005
Near Beetlegeuse
388_{10} Posts 
Thank you.

20051231, 15:26  #7 
Nov 2005
2·7·13 Posts 
Amazon has one copy of the 2nd edition if you've got $70 (new hard cover math books aren't cheap!) to expand your library. Since I'm near several libraries, colleges and universities, I think I'll be cheap and just go spend some time at a desk in one of them. This should be as interesting as the books by Howard Anton covering some of the more interesting parts of vectors and matrices  assuming you're like me and read that kind of stuff for 'fun'.
ISBN: 0387252827 (there's an older first edition #0387947779 for a little less) 
20051231, 16:35  #8 
"Nancy"
Aug 2002
Alexandria
2,467 Posts 
Some things have been added/changed in the second edition, most notably it now includes the AKS algorithm. But the first edition is still perfectly worthwhile to have and you may be able to get a second hand copy inexpensively now. Maybe check university .market newsgroups or online auctions?
Alex Last fiddled with by akruppa on 20051231 at 16:36 
20051231, 16:56  #9  
Jun 2005
Near Beetlegeuse
184_{16} Posts 
Quote:
Crandall & Pomerance, and of course Knuth were on my wish list for Christmas (again), but sadly Santa saw fit to leave me a mouse and some aftershave. Maybe if I manage to sell another picture before Easter then Amazon will be getting a call. 

20060101, 20:33  #10  
∂^{2}ω=0
Sep 2002
República de California
13·29·31 Posts 
Quote:


20060101, 22:26  #11 
Nov 2005
2×7×13 Posts 
...
Poor cat trying to eat that plastic peripheral. 
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