Author 
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SilverSprite

Posted: Fri Jun 27, 2003 2:47 pm Post subject: another quick math question for 100 


This one isnt quite as hard as darknesses question but since i wont put my bits to any good use.. here we go.. prove that 6x^2 + 2y^2 = z^2 has no natural solutions 





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AsianSensation

Posted: Fri Jun 27, 2003 3:39 pm Post subject: (No subject) 


I don't know if Im right or not(probably not), but here is mine solution:
6x^2 + 2y^2 = z^2
assuming the statement is true, i'll try to prove it by using contradiction:
first take out a common factor of 2 from the left side:
2(3x^2 + y^2) = z^2
if the above statement is true, with x,y,z belonging to Integers, then:
3x^2 + y^2 must be an odd power of 2.
let n be a number belonging to the set of positive integers
3x^2 + y^2 = 2^(2n+1)
therefore,
0 = 2^(2n+1)  3x^2  y^2
0 = 3x^2 + y^2  2^(2n+1)
0 = x^2 + y^2 + 2x^2  2^(2n+1)
0 = x^2 + y^2 + 2(x^2  2^(2n))
0 = x^2 + y^2 + 2((x  2^n)(x + 2^n))
now since x^2, y^2, and x + 2^n are all positive, therefore, for the equation to be true, x  2^n must be negative.
therefore, x  2^n < 0
x < 2^ n
sub that statement back into 3x^2 + y^2 = 2^(2n+1), and we get:
3(2^(2n)) + y^2 < 2^(2n+1)
but this statement is obviously false, because
3(2^(2n)) > 2^(2n+1)
therefore, contradiction arises, and there does not exist integer solutions to 6x^2 + 2y^2 = z^2 





SilverSprite

Posted: Fri Jun 27, 2003 4:02 pm Post subject: (No subject) 


code: 
therefore, x  2^n < 0
x < 2^ n
sub that statement back into 3x^2 + y^2 = 2^(2n+1), and we get:
3(2^(2n)) + y^2 < 2^(2n+1)

when you sub in x<2^n .. wouldnt you get 3(2^(2n)) + y^2 > 2^(2n+1) ? or am i wrong? that makes your proof false.. unless someone can tell me i'm wrong.. i hate working with these inequalities:S 





SilverSprite

Posted: Fri Jun 27, 2003 4:03 pm Post subject: (No subject) 


try again.. if you like.. anybody else? come on it isnt that hard.. 





AsianSensation

Posted: Fri Jun 27, 2003 4:05 pm Post subject: (No subject) 


am I even close, or am I like way off?
(still working on it......) 





SilverSprite

Posted: Fri Jun 27, 2003 4:08 pm Post subject: (No subject) 


i actually don't remember the solution.. i just remember doing it for night math last year.. 





SilverSprite

Posted: Fri Jun 27, 2003 4:11 pm Post subject: (No subject) 


I have realised something else.. it is false for you to assume that 3x^2 + y^2 = 2^(2n+1) 





SilverSprite

Posted: Fri Jun 27, 2003 4:17 pm Post subject: (No subject) 


yeah i just got it.. you were on the right track at the second line.. after that everything went bad 





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AsianSensation

Posted: Fri Jun 27, 2003 4:19 pm Post subject: (No subject) 


lol, i just realized that too, so instead of assuming 2^something, i'll assume it's a perfect square multiplied by 2 





SilverSprite

Posted: Fri Jun 27, 2003 4:24 pm Post subject: (No subject) 


yes that would be fair.. 





SilverSprite

Posted: Fri Jun 27, 2003 10:34 pm Post subject: (No subject) 


nobody?? come on whoa fine the new prize is all my bits.. 





Amailer

Posted: Sat Jun 28, 2003 12:35 am Post subject: IS THIS IT?: 


There is no solution because they are all different variables 





SilverSprite

Posted: Sat Jun 28, 2003 7:19 am Post subject: (No subject) 


haha..what? 





Amailer

Posted: Sat Jun 28, 2003 7:36 am Post subject: (No subject) 


Oh nothing....i just did some crap....duh im not in grd 12 but hehe i need bits! lol 





AsianSensation

Posted: Sat Jun 28, 2003 10:39 am Post subject: (No subject) 


Does this problem involves ellipse?
x^2 / (2(n^2)/ 3) + y^2 / (2(n^2)) = 1
cause when you graphed it, you get a ellipse, but how do you prove there isn't integer solutions to a ellipse? Or maybe I am on a completely different approach then what is required? (probably the latter...) 





