Pythagorean Triples

jinydu · 2003-12-05, 11:56

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.

xilman · 2003-12-05, 12:30

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

jinydu · 2003-12-06, 07:05

(3, 4, 5) and (5, 12, 13) count as separate families

(3, 4, 5) and (6, 8, 10) do not.

Wacky · 2003-12-06, 09:56

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))

rak · 2003-12-06, 13:31

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)

Drooling_Sheep · 2003-12-11, 03:04

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)

tom11784 · 2003-12-13, 10:10

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.

2003-12-05, 11:56	#1
jinydu 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.

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2003-12-06, 07:05	#3
jinydu 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.

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 6₈ 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 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.