20031205, 11:56  #1 
Dec 2003
Hopefully Near M48
2·3·293 Posts 
Pythagorean Triples
It is already known that there are an infinite number of Pythagorean Triples, for instance:
3n, 4n, 5n where n is a natural number But are there an infinite number of families of Pythagorean Triples? Prove your answer. So in the previous example: 3, 4, 5 6, 8, 10 9, 12, 15 12, 16, 20 . . . Counts as only one Pythagorean Triple family. 
20031205, 12:30  #2  
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
3×3,529 Posts 
Re: Pythagorean Triples
Quote:
Your (3,4,5) example is a primitive triple, as is (5,12,13). Only those triples (a,b,c) for which gcd(a,b,c) =1 are primitive, so (6,8,10) is not a primitive Pythagorean triple, though it is indeed a Pythagorean triple. Paul 

20031206, 07:05  #3 
Dec 2003
Hopefully Near M48
2·3·293 Posts 
(3, 4, 5) and (5, 12, 13) count as separate families
(3, 4, 5) and (6, 8, 10) do not. 
20031206, 09:56  #4  
Jun 2003
The Texas Hill Country
3^{2}×11^{2} Posts 
Quote:
(3,4,5) : k=1, n=1 (6,8,10) : k=2, n=1 (5,12,13) : k=1, n=2 from the parametric form (k*(2*n+1), k*(2*n*(n+1)), k*(2*n*(n+1)+1)) 

20031206, 13:31  #5 
Aug 2002
1 Posts 
The most interesting formula for Pythagorean Triples I know is the following:
Pick any two numbers U and V a = u^2  V^2 b = 2UV c = U^2 + V^2 A few examples: Let V = 1 and let U be an even number. u = 2 gives (3, 4, 5) u = 4 gives (15, 8, 17) u = 6 gives (35, 12, 37) u = 8 gives (63, 16, 65) 
20031211, 03:04  #6 
Aug 2003
2·3 Posts 
another one is
1)take any odd number 2)square it 3)divide by two 4)the short side will be step (1) the other two will be step(3) + .5 ex: 3^2 is 9 9/2 = 4.5 (3, 4.5.5, 4.5+.5) (3,4,5) 15^2 = 225 225/2 = 112.5 (15, 112, 113) 
20031213, 10:10  #7 
Aug 2003
Upstate NY, USA
2×163 Posts 
and it is this last formulation that allows you to prove there exist infinitely many families (well .... it gives the easiest of the formulations given)
One thing to note (making the proof even simpler) is that you have that each odd number produces a new family since the second and third differ by 1 in all cases, so they cannot be a multiple of a smaller such triple. 
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