Right angled triangle combinations(3,4,5....)

sebbysteiny · 06-10-2006

Quote:

and did you also know that in the equation of form:
a^n + b^n = c^n
has no solutions for n>2

Jay-qu

Lol, I didn't know that. I've found some solutions but not whole number solutions, which I think is what you were talking about. Nevertheless, that result is very surprising to me.

Can you show me the maths behind it?

ronthepon · 06-10-2006

Quote:

Originally Posted by sebbysteiny

Lol, I didn't know that. I've found some solutions but not whole number solutions, which I think is what you were talking about. Nevertheless, that result is very surprising to me.

Can you show me the maths behind it?

Is there a proof? I think I saw it somewhere... It was said that for 300 years since it was postulated by some french person, it has'nt been proved or disproved since.

Etc.

Janus · 06-10-2006

Quote:

Originally Posted by ronthepon

Is there a proof? I think I saw it somewhere... It was said that for 300 years since it was postulated by some french person, it has'nt been proved or disproved since.

Etc.

Fermat's last Theorum.

Fermat said he had found a proof, but never disclosed it.
A few years ago, a team using a large computer finally generated a proof of the theorum. Since Fermat did not have access to a computer, it is doubtfull that this is the same proof (if said proof ever existed).

UncleAl · 06-10-2006

Quote:

Originally Posted by Jay-qu

and did you also know that in the equation of form:
$a^n+b^n=c^n$
has no solutions for n>2

$3472073^7 + 4627011^7 = 4710868^7$

5.14880622379082621171 x 10^46

How's that?

Tim_Lou · 06-10-2006

no, thats not true, it's approximately equal, but not exact. i tried it in my calculator. i think someone proved the theorem using some kind of elliptic curve or something. its like a 200 pages+ proof. im not sure though. you might wanna check out mathworld.com

GAHD · 06-10-2006

Quote:

Originally Posted by ronthepon

Where is the math screwup?

1^2=1

so 1+1=1, thus a=1 b=1 c=1
but that is NOT a right angle triangle.

Turtle · 06-10-2006

Quote:

Originally Posted by Janus

Fermat's last Theorum.

Fermat said he had found a proof, but never disclosed it.
A few years ago, a team using a large computer finally generated a proof of the theorum. Since Fermat did not have access to a computer, it is doubtfull that this is the same proof (if said proof ever existed).

Here is a page on Wile's proof. Wile did not have a team. While Wile may have used a computer, a computer did not 'write' the proof & is not necessary for checking or understanding it. http://fermatslasttheorem.blogspot.c...ievements.html

Quote:

Originally Posted by blogspot

Wiles' proof rests on twentieth century mathematics including the theories of elliptic curves, modular forms, and Galois Representations.

Computers can be programmed to check proofs, however the possibility still exists for translation errors to & from English (Norweigen, Italian, etc.)
http://mathforum.org/kb/thread.jspa?...ssageID=284178

C1ay · 06-10-2006

Quote:

Originally Posted by GAHD

hehe, wanna see that math screw up?

Assume A and B are both 1 ...

Quote:

Originally Posted by GAHD

1^2=1

so 1+1=1, thus a=1 b=1 c=1
but that is NOT a right angle triangle.

Just because you assign A and B to one does not mean that C would be one as well. For a right triangle with both legs A and B equal to one the hypotenuse would equal the square root of 2. Math doesn't screw up.....