20091018, 08:02  #45 
Sep 2004
1000010101_{2} Posts 
I ran Gerbicz's program to search bounds less than 10,000. I got:
Number of errors=4 (too close consecutive powers) Done, time=21687 sec. So I guess there were four it couldn't tell. I don't really understand how the program works, so I don't know how to fix the errors or test just those 4 with pari or something. Also, could someone explain why for a solution if gcd(a,b) = 1 then gcd(a,b,c) = 1? And is the same true if gcd(b,c) or gcd(a,c) = 1? Last fiddled with by Joshua2 on 20091018 at 08:03 
20091018, 10:18  #46  
Jul 2006
Calgary
1A9_{16} Posts 
Quote:
Whew! That is pretty basic. If you understand GCD at all it seems it would be pretty obvious. Like GCD (1,x) = 1. You do know, I hope, that GCD stands for "Greatest Common Divisor" (or sometimes Dividend depending who you talk to). If you do GCD the slow way, get the prime factorization of each arg and list which factors are common. The most common use of GCD (and related LCM) is for arithmetic with fractions which I believe is taught in like third or fourth grade. (eg add 1/4 to 1/6). For the 3 arg analogy add 1/4 and 1/6 and 1/10. 

20091018, 14:57  #47 
"William"
May 2003
Near Grandkid
3×7×113 Posts 
Try putting it into English and pay attention to the direction of the conclusion. If p divides a and b, then it might divide a, b, and c. If there isn't any p that divides a, b, and c, then there can't be any p that divides all three.

20091018, 18:07  #48 
Sep 2004
13×41 Posts 
Sorry if the question is basic, but I'm not a math major, I'm just interested in it. I'm a computer engineer.
I understood everything lfm said, but haven't really learned anything new. I don't see how a and b not having any common factors means that neither one has any with c. Edit: I kind of get it. If something divides into a and b it must divide into c because you have a's multipled plus b's multiplied = c's multiplied and if you factor something out of a's and b's you have p(a's +b's) = c's showing that c^z is divisible by p, but is c? And for wblipp, I would put it into english, "If p divides a and b, it must divide a,b, and c and if there isn't any p that divides, a and b, no p divides into a and c or b and c." Last fiddled with by Joshua2 on 20091018 at 18:12 
20091018, 19:31  #49 
"William"
May 2003
Near Grandkid
3·7·113 Posts 

20091018, 19:38  #50  
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
1000010110111_{2} Posts 
Quote:
Here's why: 1. gcd(1,x) is 1 for any natural number x. (easy enough to see why: 1's greatest divisor is 1, and 1 divides every natural number) 2. GCD is associative, so gcd(a,b,c)=gcd(gcd(a,b),c). (I'll explain why in a bit) 3. If gcd(a,b)=1, then gcd(a,b,c)=gcd(gcd(a,b),c)=gcd(1,c)=1 (simple combination of our previously established facts) The only possible question here, then, is "why is GCD associative?" Here's why: (slightly modified from wblipp's statement) If there isn't any integer n > 1 that divides a and b, then there can't be any such n that divides a, b, and c. Quote:
(note nothing above 1 divides 10, 15, and 16, gcd(10,15,16)=1) Quote:
(note nothing above 1 divides 7, 15, and 21, so gcd(a,b,c)=gcd(7,15,21)=1) Last fiddled with by TimSorbet on 20091018 at 19:44 

20091018, 20:26  #51 
Sep 2004
13·41 Posts 
Thanks that was very informative. But we also have the condition that a^x + b^y = c^z. Does that change anything? Like you last example of a b and c didnt have any values for x y and z and it might not have worked because it wasn't a solution to that equation or something.

20091018, 20:30  #52  
Sep 2004
13×41 Posts 
Quote:


20091018, 22:45  #53  
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
11×389 Posts 
Quote:


20091019, 02:03  #54  
Jul 2006
Calgary
1A9_{16} Posts 
Quote:
GCD(2, 3, 6) is 1 since GCD(2,3) = 1 and GCD(GCD(2,3),6) is then GCD(1,6) = 1 Got it yet? Last fiddled with by lfm on 20091019 at 02:08 

20091019, 02:21  #55 
Sep 2004
13·41 Posts 
I think I get it now thanks. So if we are checking to see if we have found a counter example to the beal conjecture, all we need to check is GCD(a,b) = 1 or is there more?

Thread Tools  
Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
The Beal Conjecture Proof  Arxenar  Miscellaneous Math  1  20130907 09:59 
Distributed NFS postprocessing  poily  Msieve  6  20121205 12:45 
New Beal Conjecture Search  Joshua2  Open Projects  0  20090420 06:58 
distributed.net completes OGR25  ixfd64  Lounge  4  20081122 01:59 
distributed proofreading  adpowers  Lounge  10  20041105 01:54 