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Subject: "Triples"     Previous Topic | Next Topic
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Jonny Griffit's
guest
Jun-12-03, 02:03 PM (EST)
 
"Triples"
 
   Take the equation:

x + 8 = 2x + 3, which gives the answer x = 5

What happens if we mix up the coefficients?

2x + 8 = 3x + 1 gives x = 7
3x + 2 = x + 8 gives x = 3

It is not hard to show that three natural number solutions are the most that you can get, starting with four integer coefficients.

So {1,2,3,8} generate the Triple {3,5,7}

Pythagorean triples are generated by a formula. Can anyone find a formula to generate all these Triples, and nothing but these Triples?


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sfwc
Member since Jun-19-03
Jul-26-03, 01:06 PM (EST)
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1. "RE: Triples"
In response to message #0
 
   Unfortunately, I don't think so
but my proof it is a showcase of ugly stuff and I wasn't masochist enough to check it

Sorry

sfwc
<><


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Bractals
Member since Jun-9-03
Jul-26-03, 01:28 AM (EST)
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2. "RE: Triples"
In response to message #0
 
   Hi Jonny,

From my Number Theory class that I took many years ago - the "reduced" Pythagorean triples are given by

( u^2 - v^2 , 2uv , u^2 + v^2 ) , where

1) 0 < v < u
2) (u,v) = 1
3) u + v is odd

Hope this helps.

Bractals


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Vladimir
Member since Jun-22-03
Jul-27-03, 06:17 PM (EST)
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3. "RE: Triples"
In response to message #0
 
   LAST EDITED ON Jul-27-03 AT 10:01 PM (EST)
 
This is all a consequance of having 3 consequtive numbers and one higher number with the same parity as the middle consecutive number. Therefore, there are 2 chances of ending up with 1x and 1 chance of ending up with 2x on one side of the equation. The 1x possibilities yield integer solution by default and the 2x possibility because the higher number has the same parity as the middle consecutive number. When "mixing up" the coefficients, we have to observe the following rules:

1. 8 cannot be a coefficient of x.

Given restriction 1, there are only 6 possibilities for mixing up the coefficients to have integer solution (-7, -5, -3, 3, 5, 7).

2. Coefficient of x on the equation side with 8 must be less than the coefficient of x at the other side of the equation.

Given restriction 2, the 3 negative solutions are eliminated and we are left with only 3 possibilities - not much of a mixing up.

To generate triples (3,5,7), the 4 integers for the equation must be {(n+m)q, (n+2m)q, (n+3m)q, (n+8m)q}, n > 0, m >= 1, q >= 1 are integers. For example, n = 15, m = 2, q = 2 leads to (34, 38, 42, 62). Observing the mixing rules 1 and 2:

34x + 62 = 38x + 42 => x = 5
34x + 62 = 42x + 38 => x = 3
38x + 62 = 42x + 34 => x = 7

If you want to generate a different triple, say (2k+1, 2k+3, 2k+5), using 4 integers for the equation (n+1, n+2, n+3, p), let's see what the 4th integer p must be. The generated solutions are

2k + 1 = {p - (n + 2)}/{(n + 3) - (n + 1)} = {p - (n + 2)}/2
2k + 3 = {p - (n + 3)}/{(n + 2) - (n + 1)} = p - (n + 3)
2k + 5 = {p - (n + 1)}/{(n + 3) - (n + 2)} = p - (n + 1)

p = 4k + 4 + n from (1)
p = 2k + 5 + n from (2) or (3)

4k + 4 + n = 2k + 5 + n
k = 1
p = n + 8

The consecutive odd triples you can generate are only (3, 5, 7). But suppose you do not require the first 3 numbers (of the 4 generators) to be consecutive (or multiples of consecutive numbers) and you want to generate the triple (5, 7, 9). Denote the 4 integers 0 < a1 < a2 < a3 and p > 0.

5 = (p - a2)/(a3 - a1)
7 = (p - a3)/(a2 - a1)
9 = (p - a1)/(a3 - a2)

This gives you 3 linear equations for 3 unknowns a1, a2, a3

a1 - 9a2 + 9a3 = p
-5a1 + a2 + 5a3 = p
-7a1 +7a2 + a3 = p

Triangularization of the matrix yields

+1, -9, +9, +p
-5, +1, +5, +p
-7, +7, +1, +p

+1, -9, +9, +p
+0, -44, +50, +6p
+0, 56, +64, +8p

+1, -9, +9, +p
+0, -7, +8, +p
+0, -22, +25, +3p

+1, -9, +9, +p
+0, -7, +8, +p
+0, +0, +1, +p

Since the 3 linear equations are linearly independent, the only solution (integer or not) is the trivial solution

a3 = p
a2 = (8p - p)/7 = p
a1 = 9(p - p) + p = p

a1 = a2 = a3 = p

and you cannot generate anything with that, because the generator numbers must be different. You can do this for any desired triple. The reason you can get a generator for the triple (3, 5, 7) is that the corresponding matrix happens to have linearly dependent rows:

+1, -7, +7, +p
-3, +1, +3, +p
-5, +5, +1, +p

+1, -7, +7, +p
+0, -20, +24, +4p
+0, -30, +36, +6p

+1, -7, +7, +p
+0, -5, +6, +p
+0, -5, +6, +p

so that you can find a non-trivial solution. Should you do this for a general triple of consecutive odd numbers (2k+1, 2k+3, 2k+5), the matrix triangularization would yield a nasty cubic equation for k with integer coefficients, for which you would have to find an integer root (if it existed) and I am not going to do that. However, I am at a complete loss about the purpose of this whole exercise.


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