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Feb 23rd, 2004, 08:20 PM
#1
Thread Starter
pathfinder
Why Are these Relative Primes?
I just made the following statement in something I was documenting:
(3*c_Rows 2+1)/4 and c_Rows are relatively prime
Now, c_Rows is always odd, > 1, so 3*c_Rows2+1 is always divisible by 4.
So my question, am I right that they are relatively prime?
Can anyone prove it?
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Feb 24th, 2004, 05:31 AM
#2
Fanatic Member
hmmm
Let X = c_Rows (make it cleaner)
using the fact that gcd(a,b) <= gcd(Ya, b), and <= gcd(a^2, b)
gcd((3*X^2+1)/4, X) <= gcd(3X^2+1, X) <= gcd(3X^2+1, 3X^2)
of course, gcd(p, p+1) = 1
so, gcd(3X^2+1, 3X^2) = 1
so, gcd((3*X^2+1)/4, X) <= 1
of course, it can't be smaller, so it must = 1 !!
sql_lall 
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Feb 24th, 2004, 10:03 AM
#3
Thread Starter
pathfinder
Looks Good!
I was pretty sure it was something like that.
I'll give you proper credit in my little project.
Thanks!
-Lou
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Feb 24th, 2004, 06:51 PM
#4
Thread Starter
pathfinder
Hmmm,
I just typed the following:
With this Identity, it is easy to see that c_C and c_Rows are relative primes, that is, they share ABSOLUTELY NO COMMON DIVISOR WHATSOEVER, except for 1.
Let me Quote sql_lall when she helpfully pointed out the following:
[:clearsthroat:] Sorry for not knowing, but am I correct in referencing you as a she?
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Feb 25th, 2004, 05:34 AM
#5
Fanatic Member
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Feb 28th, 2004, 09:12 AM
#6
Fanatic Member
(3*x^2+1)/4 and x
suppose some prime p divides both numbers, then p|x (p divides x) and p|3x^2+1
but since p|x, then p|3x^2, hence p|1, so p=1
Massey RuleZ! ^-^__  Cheers!  __^-^ Massey RuleZ!
Did you know that...
The probability that a random rational number has an even denominator is 1/3 (Salamin and Gosper 1972)? This result is independently verified by me (2002)!
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Feb 28th, 2004, 01:43 PM
#7
Thread Starter
pathfinder
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Mar 10th, 2004, 09:51 PM
#8
Thread Starter
pathfinder
Hmmm,
BTW, I corrected the she to a he, sql_lall, and you are still included in my little document.
Which I just finished.
Partially titled: "Why One"
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