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Old 2009-10-18, 08:02   #45
Joshua2
 
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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 2009-10-18 at 08:03
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Old 2009-10-18, 10:18   #46
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Quote:
Originally Posted by Joshua2 View Post
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?


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.
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Old 2009-10-18, 14:57   #47
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Quote:
Originally Posted by Joshua2 View Post
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?
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.
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Old 2009-10-18, 18:07   #48
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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 2009-10-18 at 18:12
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Old 2009-10-18, 19:31   #49
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Quote:
Originally Posted by Joshua2 View Post
showing that c^z is divisible by p, but is c?
If p is prime, then yes. And that's enough.
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Old 2009-10-18, 19:38   #50
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Quote:
Originally Posted by Joshua2 View Post
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.
Nobody's saying that gcd(a,b)=1 means gcd(a,c)=1. Just that gcd(a,b)=1 means gcd(a,b,c)=1.
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:
Originally Posted by Joshua2 View Post
I would put it into english, "If p divides a and b, it must divide a,b, and c..."
This is incorrect. e.g. if a, b, and c are 10, 15, and 16 (respectively), 5 divides 10 and 15, (gcd(10,15)=5) but does not divide 16 (gcd(5,16)=1).
(note nothing above 1 divides 10, 15, and 16, gcd(10,15,16)=1)
Quote:
Originally Posted by Joshua2 View Post
I would put it into english, "... if there isn't any p that divides, a and b, no p divides into a and c or b and c."
This is also incorrect. e.g. if a, b, and c are 7, 15, and 21 (respectively), there isn't any p that divides both 7 and 15, (gcd(a,b)=gcd(7,15)=1) but 7 divides both 7 and 21, (gcd(a,c)=gcd(7,21)=7) and 3 divides both 15 and 21 (gcd(b,c)=gcd(15,21)=3.
(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 2009-10-18 at 19:44
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Old 2009-10-18, 20:26   #51
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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.
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Old 2009-10-18, 20:30   #52
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Quote:
Originally Posted by R. Gerbicz View Post
See http://robert.gerbicz.googlepages.com/beal.c ...

Note that i'm checking only gcd(a,b)=1, so it is possible that i'm printing a close hit for that gcd(a,c)>1 or gcd(b,c)>1 (obviously for a solution gcd(a,b)=1 if and only if gcd(a,b,c)=1).
This makes me think that gcd(a,c) would = 1 as well.
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Old 2009-10-18, 22:45   #53
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Quote:
Originally Posted by Joshua2 View Post
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.
I'm sure that it still works (doesn't hurt). The things I stated are true for any three natural numbers a, b, and c, regardless of why you need the GCD of those numbers. I'm not sure if its intended use in the a^x + b^y = c^z equation can be used to speed it up in some way (might help, might not help). Perhaps that's what R. Gerbicz is referring to about gcd(a,b)=1 if and only if gcd(a,b,c)=1. (in any case, gcd(a,b)=1 implies gcd(a,b,c)=1, but it's not normally also the other way around)
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Old 2009-10-19, 02:03   #54
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Quote:
Originally Posted by Joshua2 View Post
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?
Um seems you have the wrong idea of what GCD(x,y,z) is trying to do. It is looking for factors common to all three args. GCD(4, 8, 6) is just 2

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 2009-10-19 at 02:08
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Old 2009-10-19, 02:21   #55
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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?
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