Rythagorean Triples

[Martinez]

Hi, All!....my first posting here.

 

I Find the Pythagorean Triples mentioned in one of the threads and curious to know what they are....if you will, please.

[Johnny5]

Hi' date=' All!....my first posting here.

 

I Find the Pythagorean Triples mentioned in one of the threads and curious to know what they are....if you will, please.[/quote']

 

Definition: Any natural numbers A,B,C such that A^2+B^2=C^2 represent a Pythagorean triple of numbers.

 

Examples given:

 

3^2+4^2=5^2

 

6^2 + 8^2=10^2

 

So 3,4,5 is a Pythagorean triple, and 6,8,10 is a Pythagorean triple.

[Callipygous]

3,4,5

5,12,13

 

and any multiple of those, so 10, 24, 26 would be one. so is 6,8,10 and 9,12,15.

 

i think there is another commonly used set, but i forgot what it was.

[uncool]

7,24,25

8,15,17

Etc.

[Guest sandeep2357]

They are natural numbers of the form A^2+B^2=C^2

3,4,5 ETC

BUT THEY LEAD TO MORE INTRESTING PROBLEMS CAN THERE BE NUMBERS OF THE FORM A^3+B^3=C^3? OR A^4+B^4=C^4?

OR MORE GENERALLY A^n+B^n=C^n?

FERMAT STATED THAT IT IS NOT POSSIBLE TO FIND NUMBERS OF THIS SORT FOR n GREATER THAN 2. THIS IS NOW CALLED THE FERMATS LAST THEOREM. (NOT BECAUSE IT WAS THE LAST OF HIS PROPOSITIONS BUT BECAUSE IT WAS THE LAST IN ACQUIRING THE PROOF, THIS IS SO BECAUSE FERMAT WAS INTRESTED IN STATING PROPOSITIONS AND HE LEFT THE PROOF TO THE OTHER MATHEMATICIANS AS CHALLENGES.HE CLAIMED HAD PROOFS FOR THEM BUT MOST OFTEN THEY WERE NEVER FOUND. IT WAS EVENTUALLY EULER WHO SOLVED MOST OF HIS PROPOSITIONS)

[Dave]

Please don't type everything in caps - it comes across as being quite rude

[Johnny5]

Please don't type everything in caps - it comes across as being quite rude

 

I thought his keyboard was broken.

[Tom Mattson]

Here's a handy fact if you are interested in generating Pythagorean triples. Consider the Fibonacci sequence:

 

1,1,2,3,5,8,... (generate the next term by adding the two previous terms)

 

You can generate a Pythagorean triple by using the following algorithm:

 

1. Choose any 4 successive terms in the sequence.

2. Multiply the 1st and 4th.

3. Double the product of the 2nd and 3rd.

4. Add the squares of the 2nd and 3rd.

 

The results of steps 2,3, and 4 always form a Pythagorean triple.

[BigMoosie]

Thats pretty awesome Tom, I never knew such an easy method.

 

Have you heard of the pythagorean squarad (or whatever you want to call it):

 

3^3 + 4^3 + 5^3 = 6^3