Talk:Pythagorean triple/Archive 3

Pythagorean Triplet, unit circle relationships.

The treatment of the subject of the Pythagorean triple is excellent in general. I find it very instructive, but there is a minor mistake in the discussion of the unit circle relationship. The diagram showing the intersection of the line with a slope of p/q is not correctly drawn. The tangent of the angle labled as "theta", on the right-hand side of the diagram, is simply p/q, not 2pq/(p^2+q^2).

The diagram is confusing, at first one gets the impression that the triangle with its hypotenuse defined by the p/q line corresponds to the triplet triangle. This is obviously a mistake. The p/q line intersects the unit circle on the right-hand side at the coordinates of:

     x = -(q^2-p^2)/(p^2+q^2), y = +2pq/(p^2+q^2)

The hypotenuse of the triplet triangle should by defined by drawing a radial line to the above coordinates. The triplet triangle will then be illustrated by a diagram inside the left-hand positive quadrant of the unit circle.

I congratulate the author(s)of this article. I have learned a lot from it. It shows how the sides and hypotenuse of the triplet triangle can be defined in terms of p and q.

But Euclid didn't know analytical geometry, so this unit circle treatment must not have been developed until modern times. I have never read Euclid's complete works, so I am still puzzled as to how Euclid derived the p-q formulas.

Comment:

A brief, but excellent, historical discussion of Euclid's formula is given here: http://math.nmsu.edu/~history/book/euclidpt.pdf If side b = 2mn and side a = (m^2 - n^2) then tan (theta) =a/b, which must equal 2mn/(m^2-n^2), so the formula in the article is correct, but so is your formula. Note that under the "Relationships" section of the article that tan(theta/2) = n/m , which is the same as your formula, so it depends on which angle in the unit circle you're actually calculating. The following article gives an excellent summary of the various geometric relations involved in unit circles, in terms of m,n : http://www.rowan.edu/colleges/las/departments/math/facultystaff/osler/InCircle.pdf —Preceding unsigned comment added by Bronsongardner (talk • contribs) 17:14, 16 July 2008 (UTC)[reply]

Moved from Talk:Pythagorean triple/Comments

I have moved the following from Talk:Pythagorean triple/Comments. siℓℓy rabbit (talk) 14:02, 12 August 2008 (UTC)[reply]

Please find below a copy of an e-mail that I sent to Dr. Ross from www.friesian.com/pythag.htm about a simple and exhaustive method for generating triplets that is not listed in his article and also not in http: /en.wiki.x.io/wiki/Pythagorean_triple. You can use it as you see fit.

copy of e-mail:

Dear Dr Ross,

I read your article on pythagorean triplets in www.friesian.com/pythag.htm and I also read the wikipedia piece http: /en.wiki.x.io/wiki/Pythagorean_triple . I am not a mathematician, but a retired engineer, and was more or less getting into this pythagorean triplets by accident, as a friend wrote me about this and how to generate the triplets. I have come up with a way of generation that is not covered in both references above. If you have the time and deem the idea interesting, would you please comment on it. I also peruse the article with matrices in mathworld.wolfram.com, but I found it too complex for a non-mathematician.

Simple generation of Pythagorean Triplets.

If we start with the problem as in plane geometry first and start analyzing it from the start, then an exhaustive algorithm automatically emerge.

Consider the triangle ABC with sides: a opposite angle A

                                        b opposite angle B
                                        c opposite angle C, angle C being the 90 degree angle.

1. Then side c being the hypotenuse, the sides a and b will both be smaller than side c. 2. We also only need to look at the case (side a) < (side b). Should it happen that (side a) > (side b), then we flip the rectangle along the axis formed by (side a), and then let it lie on (side a) - hence making (side a) horizontal. Whence we get the case [b < a < c]

Wherefore, if a**2 + b**2 = c**2, then a**2 = c**2 - b**2 = (c+b)(c-b). We can also factor a**2, say a**2= x*y and without losing generality assume x < y. This tells us to first square a, and then to factor it in two parts x and y. This is always possible, as we can always factor a**2 = (a**2)*1. Hence we can equate:

       c+b =  y
       c-b =  x

such that the sum (x+y) = 2c =even number, and the same is true for (y-x)=2b, it also has to be even. Hence factorization where the factors are not even can be put aside. 3. The case b=0 we can forget.

Using this algorithm we can start:

   a        = 1    no solution
   a        =2    factors:     1, 4    NO solution, since sum is odd.

   a        =3    factors:    (1, 9) and (3,3). this later is trivial meaning b=0, hence no triangle.
                                   factors (1, 9) yields b=4 and c=5 or triplet [3, 4, 5]

   a        =4    factors:    (1,16), (2, 8) yielding nothing for the first one (odd sum) and [4,3, 5] the same as in a = 3

   a        =5    factors:    (1,25), (5, 5). Only for the first [5,12, 13]

   a        =6    factors:    (1,36), (2,18), (3,12), (4,9)
                       only (2,18) is of interest, yielding [6, 8, 10]

   a        =7    factors:    (1,49) yielding [7, 24, 25]
   
   a        =8    factors:    (1,64), (2,32), (4,16) with triplets [8, 63, 65], [8, 15, 17], [8, 6, 10]. Per assumption 2, we can forget this last one, as it must have been covered earlier - see case a=6
   
   a        =9        factors:    (1,81), (3,27) yielding triplets [9, 80, 82] and [9, 12, 15]

   a        =10        factors:    (1,100), (2, 50), (4, 25), (5,20)
   Only the first two yields triplets [10,99, 101] and [10, 48, 52]

and you can go on and on with integer a exhausting the set/space of all integers.

Notice that you can always factor a**2 into (1, a**2), so that you will always have a triplet [a, (a**2 - 1), (a**2+1)]. Also when a is even = 2e, then you also have a triplet [2e, 2**2{(e**2-1), (e**2+1)}] as per formula: (pa)**2 + (pb)**2 = (pc)**2

As with increasing numbers you will have many more factorization possiblities, hence more insights can be derived by looking into the factorization possibilities and its characteristics of the integers N. This should have been covered extensively in number theory (that I have not yet studied). Q.E.D.

Based on the factorization above we can easily verify the methods as given in the two references cited above.

As an example I have extended the case for a = p*q

Hence a**2 = (p*q)**2 can be factored into:

i.1 a**2 = p*(pq**2) ----> c - b = p

                               c + b = p*q**2

giving: b= [(p*q**2) - p]/2

           c = [(p*q**2) + p]/2

i.2 a**2 = (p**2)(q**2) ----> c - b = p**2

                                   c + b = q**2

giving b = [p**2 - q**2]/2

           c = [p**2 +  q**2]/2

This the solution you presented as provided by Juan Tolosa.

i.3 a**2 = (p**2.q)(q) ----> c + b = p**2.q

                                               c - b = q

giving b = [p**2.q - q]/2

           c = [p**2.q + q]/2

i.4 a**2 = (p**2 q**2) * 1 ----> c + b = p**2 q**2

                                                       c - b = 1

giving b = [p**2 q**2 -1]/2

           c = [p**2 q**2 +1]/2

Regards,

Swan-Bing HAN

bk888han@sympatico.ca, 18 Bradley Farm Court, Ottawa, Ontario, Canada K2L 4B5

--—Preceding unsigned comment added by 67.71.20.42 (talk • contribs) 17:33, June 4, 2008

Moved links from the article

I have moved the following external links out of the article. Some of these are useful and inspiring, but they do not satisfy the stringent criteria of WP:EL for inclusion in the article. Feel free to add some of these back in, with a good reason. At the moment, there are just too many links! siℓℓy rabbit (talk) 23:13, 13 August 2008 (UTC)[reply]

Formula for generating Pythagorean Triples from only one given side. How to find all possible triples from only one side of right angle triangle
http://www.math.clemson.edu/~rsimms/neat/math/pyth/ provides a Javascript calculator for the (m² − n², 2mn, m² + n²) formula, and shows how to derive the formula.
http://www.faust.fr.bw.schule.de/mhb/pythagen.htm Description of how to transform a triple into 3 new triples.
http://www.faust.fr.bw.schule.de/mhb/pythagen.htm a JavaScript calculator which illustrates the 3-fold tree structure of the set of all primitive Pythagorean triples.
Fermat's Last Theorem Blog Covers topics in the history of Fermat's Last Theorem from Pythagorean triples to Wiles' proof.
http://www.geocities.com/fredlb37/node2.html Finding m and n if given a primitive Pythagorean triple
http://www.geocities.com/fredlb37/node1.html Generating All Pythagorean Triples
http://math.nmsu.edu/~history/book/euclidpt.pdf Euclid's Classification of Pythagorean Triples
http://nrich.maths.org/askedNRICH/edited/1276.html Shows how to generate new triples by multiplication of two triples.
http://mathforum.org/dr.math/faq/faq.pythag.triples.html Formulas for Primitive Pythagorean Triples and Their Derivation
http://www.mcs.surrey.ac.uk/Personal/R.Knott/Pythag/pythag.html Links to several on-line Triple calculators; discussion of the incircle formula
http://nrich.maths.org/public/viewer.php?obj_id=1332&part=index&refpage=monthindex.php Picturing Pythagorean Triples
A Pythagorean Triple Generator and Sequencer Displays results in table form for ease-of-use.
The Online Encyclopedia of Integer Sequences contains over 160 lists associated with pythagorean triples

Requests

It currently says there are 16 triples with c < 100, but doesnt include 6,8,10...

Yes, but note that 6,8,10 is not a primitive triple. There are 16 primitive PT's with c<100. LP

I just posted the thirteenth formula for generating primitive triples. Thirteen seem like too many. Mine seems to generate all of them; it generates every primitive triple given a valid choice for the four variables. I would like someone with more knowledge to prove that it does all of them, and I would like there to be fewer than thirteen formulae, as many are redundant. Jrkenti (talk) 22:30, 12 October 2008 (UTC)[reply]

I also think we need fewer, rather than more. Unfortunately, however, we can't have any formulas that have not appeared in peer reviewed publications, per the policy against original research. I would like to remove those formula which do not generate all primitive triples. siℓℓy rabbit (talk) 22:38, 12 October 2008 (UTC)[reply]

Lem's question

Dear Silly_rabbit, I'm sorry about this, but it is much easier to edit a page than to contact someone in Wikipedia (I still don't understand your user page: no way to contact you privately.) I had basically the same problem when I tried to contact you in September. All I want -- for now -- is to know if you have an early reference for 2xy = z^2. Since I don't want to give my E-mail address away to everyone that reads this, maybe you can just post the reference in the article. The article seems to indicate that you are aware of one being needed. I will check in a few days to see if there are any changes. I have created a user page again, but know this NYC DOE address is blocked by Wiki. due to some abusers. Wiki. may again send me a confirmation to my home Juno E-mail address -- not on the internet. By the time I got this last time, I was fed up with Wikipedia, and didn't respond. Posting to my user page _might_ work -- if I can ever open it again. Again, I'm sorry for the trouble. Best wishes, "John Lemuel"

(My friend User:John Lemuel, who is new to the ways of Wikipedia, put this in the article; someone properly deleted it; I moved it here. —Tamfang (talk) 23:54, 19 October 2008 (UTC))[reply]

Now I see that Rabbit has already responded at User talk:John Lemuel; apparently Rabbit did not add the formula in question, but was merely the last to edit that section. Lem, it seems, is not the only one seeking a reference. —Tamfang (talk) 23:59, 19 October 2008 (UTC)[reply]

Matrices

It should be noted that ALL primitive triples can be generated from a given one. That is, one need not start with [3,4,5] as the article implies. The following matrix/transform, T₀, yields the predecessor of an arbitrary triple on the Barning Tree.

$\left[{\begin{array}{*{20}c}1&2&{-2}\\2&1&{-2}\\{-2}&{-2}&3\\\end{array}}\right]\bullet \left[{\begin{array}{*{20}c}a\\b\\c\\\end{array}}\right]=\left[{\begin{array}{*{20}c}{a'}\\{b'}\\{c'}\\\end{array}}\right]\to \left[{\begin{array}{*{20}c}{\left|{a'}\right|}\\{\left|{b'}\right|}\\{\left|{c'}\right|}\\\end{array}}\right]$

(Barning himself did not use this matrix). Applying the matrix/transforms T₁, T₂, T₃ finds the three "children". Iteration (in both directions) leads to the complete tree.

Note also that an entirely different ternary tree can be generated using a different set of matrices (see: http://arxiv.org/abs/0809.4324v1). —Preceding unsigned comment added by 66.66.92.105 (talk) 19:19, 16 October 2008 (UTC)[reply]

Bicorns?

I am not doing original research, but I mention bicorn in the caption to refer to the obvious arc in that first diagram. Does anyone know for a fact is those lines have a name or formula to them? Their curvature is such that my impression is that they are neither exactly upper nor exactly lower cusp, but maybe they are the average between the two cusps or something else. I am merely trying to refer to them descriptively but does anyone know of published papers or correct terminology? Also, I just arbitrarily chose the word "serendipitous". Does anyone know if a term has been coined for the triples that are the results of the generating formulas?--Intogoal2 (talk) 11:53, 24 October 2008 (UTC)[reply]

Euler's formula will generate all primitive triples, contrary to what you have suggested in your recent edit. siℓℓy rabbit (talk) 12:05, 24 October 2008 (UTC)[reply]

I was wrong and I stand corrected. I am an amateur and I apologize. I did not properly understand Euclid's formulas and now I appreciate them better and I thank you for your actions. However, I think that a simple explanation of the more-obvious radiating lines is helpful for most readers. Those bicorn-like arcs are real and they should be acknowledged. It does not have to be in the caption; it can a paragraph in the text. It is a lot of information and I think that the average reader would appreciate some guidance in order to quickly understand what they are looking at. Let me ask you another favor: do you know of any jargon relevant to the obvious patterns in that picture? I ask because, of course, the more precise and specific the jargon, the more quick Google will get me tto published papers on the matter. I will also ask a few university math profs for guidance over the next few days.--Intogoal2 (talk) 13:07, 24 October 2008 (UTC)[reply]

I have reintroduced most of my commentary because it is correct or at clearly reasonable, carefully worded and not OR, and I think you will agree that it is now less intrusive.--Intogoal2 (talk) 14:12, 24 October 2008 (UTC)[reply]

David Eppstein: Please let us collaborate and not just bully each other. What I have added is factual as matches the triples presented and thus not OR. I known those arcs are probably not bicorn, but comparing them as similar to bicorns helps the reader to see what I am talking about. How do you propose we describe those patterns such that we would all quickly recognize what is being referred to?--Intogoal2 (talk) 15:33, 24 October 2008 (UTC)[reply]

It is not enough for these patterns to be factual. They must also be attested in the literature. —David Eppstein (talk) 17:02, 24 October 2008 (UTC)[reply]

How is it helpful to say "There are arcs that look to me like bicorns"? If you can see the arcs, so can another reader. The proper thing to do, imho, would have been to ask here, "What are these arcs that I see? Are they bicorns or something else? Shouldn't the article say something about them?". —Tamfang (talk) 03:36, 25 October 2008 (UTC)[reply]

Pretty good for rank amateur. Huh? I am feeling a little less silly, Silly Rabbit, because you are clearly a mathematician of some sort.--Intogoal2 (talk) 16:38, 24 October 2008 (UTC)[reply]

I'd rather have this as a discussion of the density of Pythagorean triples. Awhile ago, I was considering adding a section on ergodicity of the Barning process essentially due to Dan Romik (2004), and this in turn ties in with "chaos theory" (that is, dynamical systems). It would seem to be better to have a full section on density properties rather than a sophomoric section entitled "On the scatter plot". Compared to Euclid's formula, the scatter plot is of rather marginal significance. I'll move the section to the end of the article with improvements later today. siℓℓy rabbit (talk) 16:49, 24 October 2008 (UTC)[reply]

Look, I just saw those curves and figured since smaller versions of those plots have been affordable for about 30 years now, somebody noticed them, did the analytical work and blah blah blah. I am groping and wording is amateurish but we now have some quality references. It is an Age of Computers thing, much like Bailey–Borwein–Plouffe formula. It was always there and it could have been found analytically, but a computer helped us to recognize it (well, to plot the triples and then notice the patterns, whereas with BBP, it could have been discovered analytically but, in fact, a computer search found it first). Perhaps it is best quoted from The Quest for Pi:

However, this derivation (provided in the preceding paragraphs) is dishonest, in the sense that the actual route of discovery was much different. This formula was actually discovered not by formal reasoning, but instead by numerical searches on a computer using the “PSLQ” integer relation finding algorithm. Only afterwards was a rigorous proof found.

But please guide the naive user with descriptive language and referring back o the diagram so that s/he can easily see what we are talking about before diving into the equations and less-accessible theory. For the high schooler and even grade schooler who can just as easily see those slightly scattered curves.

SR: I am sure that the BBP article(s) would also benefit from your skillful editorial hand.--Intogoal2 (talk) 17:22, 24 October 2008 (UTC)[reply]

Aha! They are all parabolas with all of their foci at the origin. Now I know. Done!--Intogoal2 (talk) 17:43, 24 October 2008 (UTC)[reply]

Cleanup proposal

I propose to remove generating formulas that lack citation in the near future. My reason for doing so is the following. Originally, the section read that these formulas generate all Pythagorean triples. This is true for the first few formulas, but I haven't checked the others (and I shouldn't have to check them, there should be a citation). The article should, in my opinion, clearly indicate which formulas generate all (or at least all primitive) triples, and which only generate some triples. In fact, I'm not sure whether we should cover formulas that only generate some triples: surely such a formula must be notable in some way in order to deserve coverage here. This is another thing that citations will help to resolve. We now have a glut of formulas, and it is obviously misleading to some readers, since there is a view that not every triple can be generated by the formulas in the article. siℓℓy rabbit (talk) 12:20, 24 October 2008 (UTC)[reply]

Reply to Cleanup Proposal

Rabbit proposes —Preceding unsigned comment added by74.170.183.119 (talk) 14:30, 26 October 2008 (UTC) 74.170.183.119 (talk) 13:00, 26 October 2008 (UTC)[reply]

Citations are still needed. In fact, they are preferable to proofs. See WP:V for more details. siℓℓy rabbit (talk) 13:16, 26 October 2008 (UTC)[reply]

Citations can be found for the incompossible claims that FDR was one of our greatest presidents, and that he was one of the worst. Such claims, even with citations, are contentious, and Wikipedia probably should not endorse either. I have been unable to find any source for the matrix M in section VI, or the equations for a(r,k), b(r,k) and c(r,k) that generate it, but those equations and that matrix are surely not contentious, and I have presented valid algebraic proof that they are as described. That knowledge should be preserved for posterity, even if no one else before me discovered these facts. Somebody probably did, but I was unable to find any published source. I am 73 years old, retired, and credit for the discovery will not benefit me, even in the unlikely event that I am entitled to it, but the matrix M makes clear the existence of an infinite number of families of PNTs, not just the two families produced by the well-known formulae of Plato and Pythagoras, and the relationships between them. This knowledge should be available to interested persons, and especially to teachers. I do not propose to get into an editing war with anybody; you can, of course, delete anything you choose to, but I submit that humanity will be better served if section VI remains.74.170.183.119 (talk) 14:30, 26 October 2008 (UTC)Bret Hooper, 615-217-8578, brethooper@gmail.com[reply]

The article already contains much more original research than I am comfortable with. Regardless of what is good or bad for humanity, the article will be better served by enforcing a strict policy against WP:OR. I propose that the section under discussion (section VI) should be rewritten from the sources provided for it. Currently, the only source in this section is Martin (1875), which only describes the construction of Pythagorean triples from the Pell numbers. siℓℓy rabbit (talk) 14:46, 26 October 2008 (UTC)[reply]