20120410, 07:44  #1  
Basketry That Evening!
"Bunslow the Bold"
Jun 2011
40<A<43 89<O<88
1C35_{16} Posts 
Classes
While I'm waiting for that polyselect to finish, I thought I'd post my provisional (read: desired) schedule for next semester (FA12). There's probs a less than 50% chance it will actually be this, but anyways:
MATH 453: Elementary Theory of Numbers Quote:
Quote:
Quote:
Quote:
Quote:
Last fiddled with by Dubslow on 20120410 at 07:45 

20120912, 05:17  #2 
Basketry That Evening!
"Bunslow the Bold"
Jun 2011
40<A<43 89<O<88
3·29·83 Posts 
Heh, here's question one of my number theory hw due tomorrow:
26. Prove or disprove the following statements. a) If p is a prime number, then 2^p1 is a prime number. b) If 2^p1 is prime, then p is prime. Last week I had to prove that if a^n1 is prime, n is prime and a==2. 
20120912, 06:13  #3 
Romulan Interpreter
"name field"
Jun 2011
Thailand
2×3×31×53 Posts 
Haha, that is nice!
But from your former posts I understood (and I was totally convinced) that you go to some college/university, not to kindergarten... 
20120912, 09:11  #4 
Basketry That Evening!
"Bunslow the Bold"
Jun 2011
40<A<43 89<O<88
3·29·83 Posts 
Hey, that's only two weeks of class, give us time... (I'm halfhoping we'll be able to get into the workings of NFS, but I'm rather doubtful, especially that our scope is broad/deep enough for NFS.)

20120912, 10:13  #5  
Noodles
"Mr. Tuch"
Dec 2007
Chennai, India
2351_{8} Posts 
Quote:
Γ: 2 is a primitive root (mod p) Λ: 2 is a quadratic nonresidue (mod p) Σ: p ≡ 3, 5 (mod 8) Ω: p is NOT a prime factor of some Mersenne number 2^{q}1, with q < p1 For these four things, you can be able to be, being include the NOT clause for these following four statements, given p is being a prime ≥ 2. Π: p ≡ 3 (mod 4) Ψ: p is the sum of two squares Θ: 2p+1 is prime, with 2p+1  2^{p}1 Φ: 2p+1 is prime, with 2p+1  2^{p}+1 By the way, this was just simply being a fun problem I gave, don't mind it taking it, solving it very seriously, during the time periods. Last fiddled with by Raman on 20120912 at 10:16 

20120912, 22:41  #6  
Basketry That Evening!
"Bunslow the Bold"
Jun 2011
40<A<43 89<O<88
3·29·83 Posts 
This really belongs in the Puzzles forum, but here goes...
Another of the problems from the homework I turned in today was the following: Quote:
 ab and ab > amb+nc for some m,n in Z;  if a=qb+r for a given a,bq,r in Z, then 0<=r<b, and q and r are unique;  we've defined GCD(a,b), and  proven that GCD(a,b) = min{ma+nb  m,n in Z, ma+nb>0}, i.e. that GCD(a,b) is the smallest positive integer that's a linear integral combination of a and b (and in particular, such a combination exists). The (semiconstructive) proof I came up with at 3am only depended on what we've covered, except that I needed Chebyshev's theorem. (I can reproduce it if anyone desires.) Is there a proof of the statement without relying on that theorem? 

20120912, 23:06  #7  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
Quote:
Last fiddled with by science_man_88 on 20120912 at 23:09 

20120912, 23:29  #8  
∂^{2}ω=0
Sep 2002
República de California
13·29·31 Posts 
Quote:
1. If n odd and n > 3, n = 2 + k with k odd and k > 1, hence 2,k coprime; 2. If n even and n > 4 with n == 0 (mod 4), we have n = (n/21) + (n/2+1) with each summand odd and > 1. Since the 2 summands differ by 2, they cannot share any odd factors; 3. If n even and n > 6 with n == 2 (mod 4), we have n = (n/22) + (n/2+2) with each summand odd and > 1. Since the 2 summands differ by 4, they cannot share any odd factors. 

20120912, 23:43  #9  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
Quote:
odd1=even+odd2=floor(odd1/2)+ceiling(odd1/2) these 2 values have a difference of one. since for a>b, GCD(a,b) must divide ab and ab = 1 GCD(a,b)==1 hence they are coprime. 

20120912, 23:56  #10  
Basketry That Evening!
"Bunslow the Bold"
Jun 2011
40<A<43 89<O<88
3×29×83 Posts 
Quote:
Last fiddled with by Dubslow on 20120912 at 23:59 

20120913, 00:11  #11  
"Forget I exist"
Jul 2009
Dumbassville
10000011000000_{2} Posts 
Quote:
also: Quote:


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