 mersenneforum.org Pythagorean Triples
 Register FAQ Search Today's Posts Mark Forums Read 2003-12-05, 11:56 #1 jinydu   Dec 2003 Hopefully Near M48 6DE16 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.   2003-12-05, 12:30   #2
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3·43·79 Posts Re: Pythagorean Triples

Quote:
 Originally posted by jinydu But are there an infinite number of families of Pythagorean Triples? Prove your answer. So in the previous example:
Yes, or no. It depends on whether the moderately well-known parameterization of ALL primitive Pythagorean triples is regarded as a single family.

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   2003-12-06, 07:05 #3 jinydu   Dec 2003 Hopefully Near M48 6DE16 Posts (3, 4, 5) and (5, 12, 13) count as separate families (3, 4, 5) and (6, 8, 10) do not.   2003-12-06, 09:56   #4
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Jun 2003
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43A16 Posts Quote:
 Originally posted by jinydu (3, 4, 5) and (5, 12, 13) count as separate families (3, 4, 5) and (6, 8, 10) do not.
Then, as Paul noted, there are infinitely many "families" since you have chosen to define "family" in that manner.

(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))   2003-12-06, 13:31 #5 rak   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)   2003-12-11, 03:04 #6 Drooling_Sheep   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)   2003-12-13, 10:10 #7 tom11784   Aug 2003 Upstate NY, USA 1010001102 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.  Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post Alberico Lepore Alberico Lepore 43 2018-01-17 15:55 a nicol Miscellaneous Math 21 2017-12-19 11:34 Nick Number Theory Discussion Group 2 2016-12-18 14:49 bhelmes Math 2 2016-07-20 10:34 Rokas Math 3 2005-01-02 03:50

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