
Re: Pythagorean triples
Posted:
Mar 13, 2013 3:45 AM


> There are some Pythagorean triples ( = integers > {a,b,c}: a^2+b^2c^2=0) such that the shorter two > sides differ by only 1. E.g. 20, 21, 29 ; 119, 120, > 169. > > Is there a finite number of such triples? If so, how > many? > > Or > > Show that there is an infinite number.
From: Doctor Nisith Bairagi. My email address : bairagi605@yahoo.co.in Date : March 13, 2013 Subject: SPECIAL PYTHAGOREAN TRIPLES: GENERAL FORMULAE Re: Pythagorean Triples(posted : June 26, 2006)
Dear Mathematicians, I have read the interesting question posed by Jason Osborne (07/10/97) and subsequent discussion by Doctor Rob (07/11/97) on the special type of Pythagorean triples of the type (119,120,169), (696,697,985), ?, in which the two adjacent sides of a right triangle forming right angles, are consecutive numbers, (i.e., x and y differ by 1), while z is the hypotenuse, satisfying x^2 + y^2 = z^2.
I send a straight forward and valid answer to this quarry (omitting the derivation part) as follows:
By putting: x = (u^2 ? v^2) = (u + v)(u v), y = 2uv, and z = (u^2 + v^2), and putting numerical values for u and v, the infinite sets of all Pythagorean triples (including the special type in question), can be written down directly.
Here, we present this special category of Pythagorean triples in this form: (u,v : x, y, z) through (1) the sequences of (u, v, and z), and also through (2) the proposed formulae, as follows:
(1) Proposed Sequence: u sequence : 1, 2, 5,12, 29, 70, 169, 408, 985, 2378, ?(for n = 1,2,3?.). [start with u(1) = 1, [u(n) = 2u(n1) + u(n2)], (169 = 2 x 70 + 29)] v sequence : 2, 5,12, 29, 70, 169, 408, 985, 2378, 5741, ?(for n = 1,2,3?.). [start with v(1) = 2, from the same usequence, [(v(n) = u((n+1)), (v(8) = u(9) = 2378)]. The xsequence and ysequence can be easily calculated from the u and vsequence. It will be noted that all the numbers of the xsequence are odd and factorable. Again, if the nth term is even, x >y, we get x  y =1, and if odd, y >x, we get y ? x =1. This is controlled by : /x ? y/ = (1)^n, for any even or odd value of n.
z sequence : 5, 29, 169, 985, 5741, 33461, 195025, ?(for n = 1,2,3?.). [start with z (1) = 5, and dropping alternate terms in usequence, [z(n) = 6z(n1) z(n2), (5741 = 6x 985  169)], Also: [(z(n) = v(2n) = u(2n+1)), (z(5) = v(10) = u(11) = 5741)].
(2) Proposed Formula: Without resorting to the sequence form, the terms u(n), v(n), and z(n) can be directly calculated as:
v(n) = [((2)^(0.5) +1)^n)/(2(2)^(0.5)) + (1)^(n1).((2)^(0.5) 1)^n)/(2(2)^(0.5))] u(n) = v(n1)
z(n) = [(5(2)^(0.5)7).((32(2)^(0.5))^(n1)) / (2(2)^(0.5)) + (1)^(n1).(5(2)^(0.5)+7).((32(2)^(0.5))^(n1)) / (2(2)^(0.5))].
Thus, for n = 9, the 9th triple is: (u, v: x, y, z) = (2378, 985, 4684659, 4684660, 6625109).
For larger and larger terms, the ratio B = z(n+1)/z(n) converges to [3+2(2)^0.5], or, its reciprocal 1/B = z(n)/z(n+1)to [32(2)^0.5]. This ratio number B (or its reciprocal 1/B), enjoys the property that: (B + 1/B) = 6. Compare this property with that of Fibonacci sequence of numbers: 1, 1, 2, 3, 5, 8, 13, 21, 34, ?,where the difference of F = F(n+1)/F(n) = [(1 + (5)^0.5] / 2, and its reciprocal 1/F = F(n)/F(n+1)= [(1 + (5)^0.5] / 2, yields: (F 1/F) = 1.
All the infinite sets of special Pythagorean triples, in which out of the three, the two are consecutive numbers, can be obtained easily either by (1) the proposed sequences, or, by (2) the direct application of the formulae, as shown here.
[For further details, refer to the Book ?Advanced Trigonometric Relations through Nbic Functions? by Nisith K Bairagi, New Age International Publishers, New delhi (2012), Appendix A].
Does this solution satisfy your required enquiries? Please acknowledge/accept my proposed solution, and produce this solution in your column for other readers, and please reply/communicate your comments to me in my Email.
From: Doctor Nisith Kumar Bairagi My email: <bairagi605@yahoo.co.in>
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