another quick math question for 100

[AsianSensation]

that's it, I give up, I can't find a proof to this......

it seems like a simple problem, but I just can't seem to do it.....damn, my math skills is getting rusty....

[SilverSprite]

Quote:

damn, my math skills is getting rusty....

so are your english skills...

well anyways your way off... it has nothing to do with ellipses.. nobody else seems to be trying it though.. oh wells i guess i get to keep my bits then

[AsianSensation]

post the solution, or rather, if you don't want to post it, PM me, I want to see what I did wrong, or what I didn;t do.

[Crono]

aights, since this is a comp sci forum, i guess u guyz could try n use turin 2 solve tis problem, it'll b real easy if u kno the approach, but i guess the math way is "cool" 2...haha

[bugzpodder]

my turn: 6x^2 + 2y^2 = z^2
assuming (x,y,z) is a primitive solution (ie gcd(x,y,z)=1). otherwise you can take out a common factor gcd(x,y,z)^2 and so forth

6x^2=z^2-2y^2

in mod 6

0=z^2-2y^2 (mod 6)
so
z^2=2y^2 (mod 6)

if y=0 (mod 6), z^2=0 (mod 6)
y=1(mod 6), z^2=2 (mod 6)
y=2 .., z^2=2 (mod 6)
y=3..., z^2=0 (mod 6)
y=4..., z^2=2 (mod 6)
y=5..., z^2=2 (mod 6)

you can verify that there is no solution to z^2=2 (mod 6)
and of course, z^2=0 (mod 6) implies that z=6z' for some integer z' and y=3y'

6x^2=36(z')^2+18(y')^2
x^2=6(z')^2+3(y')^2
and it follows that x=3x'
but gcd(x,y,z)=gcd(3x',3y',6z')>=3 ... we assumed it to be 1. contradiction

[Andy]

wow go bugz..

[SilverSprite]

dodge_tomahawk wrote:

wow go bugz..

haha that means you dont understand a word he said lol

[Andy]

shut up shut up... that means he's too good that i dun understand what he said

[SilverSprite]

dodge_tomahawk wrote:

shut up shut up... that means he's too good that i dun understand what he said

nono it just means that you dont know what he said nothing more.. i'll be quoting bugz

bugzpodder wrote:

..its common sense..

[bugzpodder]

so do i get the 100 bits? ^.^' or is that just like 100 in jeopardy?

[eggplant_burger]

instead of using my math skillz (which i don't really have, my only 12 course is calculus, so far) to solve this rpoblem, I have decided to use my programming skillz. Here it is:

var doneyet : boolean := false
for x : 1 .. maxnat
for y : 1 .. maxnat
for z : 1 .. maxnat
if 6 * (x ** 2) + 2 * (y ** 2) = z ** 2 then
doneyet := true
put "Eureka!!", skip, "I have found it", skip, skip,
"when x=", x, " y=", y, " z=", z, " then it is true",
skip, skip, "MUWAHAHAHAHAHAHA"
end if
end for
end for
end for
if doneyet = false then
put "Wow, there really ISN'T a natural solution... Trippy"
end if

That'll prove to you that there is no natural solution. And it was fun to write.

[bugzpodder]

the limitations of maxnat is only a computerized limitation. there is no limiations to natural numbers, and therefore, unfortunately, it is not a valid proof

[eggplant_burger]

that's a piddling argument. it can be proven that it is a valid proof.

since there are 2 variables to a degree greater than 1 on the left side, and only 1 variable to a degree greater than 1 on the right side, the left will increase more quickly than the right. So if they are already not equal at a point where the values of the left and right side fall into a pattern, they will never be equal. I think that the maximum possible natural number in Turing (4294967294) is reasonably high enough to assume that a pattern will be established. It's like using limits... you don't quite show it exactly, but it's so close it doesn't make a difference.

[PaddyLong]

no... it's not a propper proof... it's only showing it for n number of cases rather than for all possible

[SilverSprite]

hmm i think he was just kidding?