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

[Roadam]

We all probably know that combination. Is there any other combination than this to fit into a^2+b^2=c^2? Other than its multiples(6,8,10) of course.

[cwes99_03]

We all probably know that combination. Is there any other combination than this to fit into a^2+b^2=c^2? Other than its multiples(6,8,10) of course.

You are asking about square roots. Forget the whole triangle (thought the question remains the same whether you think of it as a triangle or not).

 

You are asking if the sum of the squares of any two numbers, equals the square of another number.

 

Do some number crunching, see if you can find any combinations. 25^2-24^2-=?^2.

[cwes99_03]

Wow, just pulled that out of my . Must have seen it before.

[GAHD]

hehe, wanna see that math screw up?

 

Assume A and B are both 1 ...

[TheBigDog]

We all probably know that combination. Is there any other combination than this to fit into a^2+b^2=c^2? Other than its multiples(6,8,10) of course.

Try this on for size. There is also some discussion in the Katabatak thread. Or look up primitive right triangles on Wiki. There are an infinite number of unique integer right triangles.

 

Bill

[UncleAl]

Draw a circle on a piece of graph paper with its center at the origin. Every point on the circle satisfies the relationship x^2 + y^2 = (radius)^2. Pick a radius, pick a point, then read off its coordinates.

[Tim_Lou]

these triples are in the forms of:

[math]a=2xy[/math]

[math]b=x^2-y^2[/math]

[math]c=x^2+y^2[/math]

using these equations, you can generate these triples:

example:

let x=7, y=5

a=2*5*7=70

b=49-25=24

c=49+25=74

 

[math]24^2+70^2=74^2[/math]

 

you can prove it yourself:

[math]a^2+b^2=(2xy)^2+(x^2-y^2)^2=c^2[/math]

[Roadam]

Wow, amazing.

Looks like I have lots of math yet to learn.

[Jay-qu]

and did you also know that in the equation of form:

[math]a^n+b^n=c^n[/math]

has no solutions for n>2

[ronthepon]

hehe, wanna see that math screw up?

 

Assume A and B are both 1 ...

Where is the math screwup?

[sebbysteiny]

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]

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]

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]

and did you also know that in the equation of form:

[math]a^n+b^n=c^n[/math]

has no solutions for n>2

 

[math]3472073^7 + 4627011^7 = 4710868^7[/math]

 

5.14880622379082621171 x 10^46

 

How's that?

[Tim_Lou]

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]

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]

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.com/2005/05/fermats-achievements.html

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?threadID=66901&messageID=284178