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another quick math question for 100       Author Message
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    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    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...) Display posts from previous: All Posts1 Day7 Days2 Weeks1 Month3 Months6 Months1 Year Oldest FirstNewest First        Page 1 of 3  [ 34 Posts ]
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