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Old 2003-12-05, 11:56   #1
jinydu
 
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Default 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.
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Old 2003-12-05, 12:30   #2
xilman
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Default 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
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Old 2003-12-06, 07:05   #3
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(3, 4, 5) and (5, 12, 13) count as separate families

(3, 4, 5) and (6, 8, 10) do not.
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Old 2003-12-06, 09:56   #4
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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))
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Old 2003-12-06, 13:31   #5
rak
 
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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)
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Old 2003-12-11, 03:04   #6
Drooling_Sheep
 
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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)
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Old 2003-12-13, 10:10   #7
tom11784
 
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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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