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Flofloflo
Member
Joined: 07 Nov 2007 Posts: 120

Posted: 01 Jul 2009 06:24:11 am Post subject: 


Hello,
I am playing with the Collatz Conjecture for a bit, and I got to a problem.
Let's say you have an equation with two variables; 16Y2 = 3Z
I need to find the formula for all whole numbers Y for whom there exists a whole number Z so that 16Y2 = 3Z.
One way to do this would be to write down a table with both formulas and look for the common numbers and make a formula for these numbers, but this takes too much time.
I reckon this might be a really easy problem but I just can't find out how to do it.
Can somebody help maybe?
Floris 

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thornahawk μολών λαβέ
Active Member
Joined: 27 Mar 2005 Posts: 569

Posted: 01 Jul 2009 09:16:39 am Post subject: 


What you're looking for is a method to solve a linear Diophantine equation. There's a number of them, go pick.
(edit:
Okay, I tried your equation on Mathematica; the general solution is y==3n1 and z=16n6, for all integer values of n.
The particular Mathematica command was [font="Courier New"]Reduce[16y  2 == 3z, {y, z}, Integers]
)
thornahawk
Last edited by Guest on 01 Jul 2009 09:30:29 am; edited 1 time in total 

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Flofloflo
Member
Joined: 07 Nov 2007 Posts: 120

Posted: 01 Jul 2009 11:12:39 am Post subject: 


Hey thanks for the info.
I tried some stuff though, and it seems to me just using the table of the calculator to find the first solution is way easier then using Euclids method...
And then I can just use:
ax + by = c
a(x0) + b(y0) = c
(use euclid or table of maybe there's an easier way? to find X0)
x = x0 + bt
Edit, now that I think about it, I am gonna let my TI do this anyways so actually it doesn't matter if Euclid's method is slow if you do it by hand...
edit & of course I just need to take the bmodulo of X0 .
Edit, forget it, I am just gonna use wolfram .
But seriously I feel like I am solving the Collatz Conjecture
Last edited by Guest on 01 Jul 2009 12:38:56 pm; edited 1 time in total 

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Flofloflo
Member
Joined: 07 Nov 2007 Posts: 120

Posted: 04 Jul 2009 08:15:09 am Post subject: 


Hello,
I am pretty sure I've gisted down the collatz conjecture to one problem, that seems pretty solvable:
a1,a2,a3...an and b1,b2,b3....bn are positive integers. a1,a2... an are all different, the same goes for b1, b2.... bn
Consider all permutations of 2^(a1)*3^(b1) + 2^(a2)*3^(b2) ..... + 2^(an) * 3^(bn)
If there's a possibility of getting two different permutations that yield the same result, the Collatz conjecture isn't verified, but not necesarily falsified. If this is impossible the Collatz conjecture is verified (I think).
So if somebody can help me with this I guess I'll be all done with the whole collatz thing; if this doesn't work I'll give it up, but this seriously should work.
edit, crap, maybe I actually need to know something else.
I might need to know wether, with a Linear Diophantine Equaton with infinte solutions, of the form ax+by = c, different values a, b and c can result in the same solutions.
Last edited by Guest on 04 Jul 2009 11:00:12 am; edited 1 time in total 

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thornahawk μολών λαβέ
Active Member
Joined: 27 Mar 2005 Posts: 569

Posted: 04 Jul 2009 11:11:35 am Post subject: 


I had almost forgotten that I had already posted an extended GCD program that might interest you.
As for
Quote: with a Linear Diophantine Equation with infinte solutions, of the form ax+by = c, different values a, b and c can result in the same solutions.
well obviously, you can multiply a, b, and c with an integer n and you'd still have the same solutions x and y, assuming they exist (the conditions for existence are in a number theory textbook).
thornahawk 

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Flofloflo
Member
Joined: 07 Nov 2007 Posts: 120

Posted: 04 Jul 2009 11:40:45 am Post subject: 


okay, so that's the only way to get the same solutions I suppose.
pff, I will be studying math next year... I'm looking forward to it already .
But anyways... I'm going to see if I can formulate my evidence that the collatz conjecture is true. Altough I might need the other thing I asked for that, the whole 2^A1*3^B1... thing.
Btw, gonna check out the euclid program you wrote to.
Last edited by Guest on 04 Jul 2009 11:41:35 am; edited 1 time in total 

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DarkerLine ceci n'est pas une 
Super Elite (Last Title)
Joined: 04 Nov 2003 Posts: 8328

Posted: 04 Jul 2009 02:14:47 pm Post subject: 


Unfortunately, your condition seems a little too strong. Here is a counterexample:
2^{8}*3^{3} + 2^{2}*3^{0} + 2^{0}*3^{8} + 2^{1}*3^{2} = 13495
2^{2}*3^{3} + 2^{8}*3^{0} + 2^{1}*3^{8} + 2^{0}*3^{2} = 13495
(a1,a2,a3,a4)=(8,2,0,1) are all different, so are (b1,b2,b3,b4)=(3,0,8,2).
If you prefer strictly positive integers, feel free to multiply through by 6, which will add 1 to every exponent.
Last edited by Guest on 04 Jul 2009 02:16:29 pm; edited 1 time in total 

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Flofloflo
Member
Joined: 07 Nov 2007 Posts: 120

Posted: 05 Jul 2009 06:53:26 am Post subject: 


Aaargh... Well, it was a pretty bold claim I guess. It'll probably a bit more complicated then I thought to prove the conjecture . 

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