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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.wikipedia.org/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.wikipedia.org/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 a2 + b2 = c2, then a2 = c2 - b2 = (c+b)(c-b). We can also factor a2, say a2= xy 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 a2 = (a2)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 a2 into (1, a2), so that you will always have a triplet [a, (a2 - 1), (a2+1)]. Also when a is even = 2e, then you also have a triplet [2e, 22{(e2-1), (e2+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 = pq Hence a2 = (pq)2 can be factored into: i.1 a2 = p(pq2) ----> c - b = p c + b = pq*2 giving: b= [(pq*2) - p]/2 c = [(pq2) + p]/2 i.2 a2 = (p2)(q2) ----> c - b = p2 c + b = q2 giving b = [p2 - q2]/2 c = [p2 + q2]/2 This the solution you presented as provided by Juan Tolosa. i.3 a2 = (p2.q)(q) ----> c + b = p2.q c - b = q giving b = [p2.q - q]/2 c = [p2.q + q]/2 i.4 a2 = (p2 q2) * 1 ----> c + b = p2 q2 c - b = 1 giving b = [p2 q2 -1]/2 c = [p2 q2 +1]/2 Regards, Swan-Bing HAN bk888han@sympatico.ca, 18 Bradley Farm Court, Ottawa, Ontario, Canada K2L 4B5

This is the talk page for discussing improvements to the Pythagorean triple article.
Please sign and date your posts by typing four tildes (`~~~~`). Put new text under old text. Click here to start a new topic. New to Wikipedia? Welcome! Ask questions, get answers. This is not a forum for general discussion about the article's subject.	Be polite Assume good faith No personal attacks Be welcoming	Article policies No original research Neutral point of view Verifiability

1 Property of Triples
2 Requests
3 God Created the Integers
4 Pythagorean triple
5 mis-wording?
6 Prime numbers??
7 A Pythagorean Triple which approximates 30,60,90
8 Pythagoras' Theorem and the Platonic Sequence: A Common Genesis
9 "Below is a formula created by Matthew Irvine"
10 Diophantus II.VIII and the Platonic Sequence
11 Pythagorean Triplet, unit circle relationships.
12 Suggested links

[edit] Property of Triples

Mr. Parker's reply to my comments was excellent--and instructive. I have tried to summarize his remarks and added that summary to the "Platonic Sequence" section. Bronson Gardner

Thankyou for your kind words but honour enough to have contributed a sentence or two on a topic which is of such great significance in the history of Maths. In the interests of possible further improvements on the page, may I humbly suggest that some careful thought be given to the slight contradiction posed by this contribution: namely that on the one hand the Platonic sequence is said to be a 'special case' of the general formula (m^2-n^2,2mn,m^2+n^2) with n=1 whilst on the other it can be shown that the 'general formula' itself may be derived from the Platonic sequence ([t^2-1]/2;t;[t^2+1]/2) when the parameter t is a rational number m/n. If you feel we are now in the realms of general discussion, please feel free to shift this paragraph to my user page and if needs be, take it from there. Neil Parker 16:08, 5 July 2007 (UTC)

I removed this statement from the "Properties of Triples Section": "The smallest number of the triple is always the square root of the sum hypotenuse and the other leg." The statement is partly true. There are some triples in which the smallest side is equal to the sqr (hypotenuse + largest side) (examples:3,4,5 and 5,12,13) But, the statement is not generally true, since most triples that do not satisfy this formula. Examples: For the triples 6,8,10 and 8,15,17 , the sqr (8+10) does not equal 6 and the sqr(15+17) does not equal 8. In reviewing a table showing the first 365 triples, 344 triples did not fit this formula, but 21 triples did satisfy the formula.

—The preceding unsigned comment was added by Bronsongardner (talk • contribs) 17:14, 26 March 2007 (UTC).

Actually - with some conditionals - your statement is always true if we accept that all Pythagorean triples are derivatives of the Platonic sequence (see my note below). Taking (x,y,r) as (6,8,10) and (8,15,17) in your examples, divide through by r-x with t = y/(r-x) = 2 in the first case and 5/3 in the second. Now you have equivalent rational triples which satisfy the Platonic sequence (x,y,z) = ([t^2-1]/2 ; t ; [t^2+1]/2) and therefore the property you have perhaps over hastily removed! The problem is with the somewhat arbitrary definition of Pythagorean triples which insists they must specifically be integers rather than just rationals. So the Platonic equivalent of (6,8,10) which is (3/2; 2, 5/2) is unfortunately 'disqualified' from the realm of Pythagorean triples! We also need to note the condition that if the calculated t-value is less than (1+sqrt2), the middle side (not the smallest side) will be the square root of the sum of the other two. This will be true for the Platonic equivalents of both your given 'counter' examples. Of course we could always swap the legs and write (8,6,10) with t-value 3 and Platonic equivalent (4,3,5) satisfying your statement completely. Also (15,8,17) with t-value 4 and Platonic equivalent (15/2; 4 ; 17/2). Neil Parker 17:42, 17 April 2007 (UTC)

ALL Pythagorean triples are derivatives of the basic Platonic sequence t, (t^2-1)/2 and (t^2+1)/2. So if we want to generate any triple this is all we need. It follows that every triple has a corresponding rational t value which can be used to generate a similar (ie equiangular) triangle with rational sides in the same proportion as the original. In fact if you replace t with a rational fraction m/n in the sequence , you will arrive at the 'standard' triple generator. 2mn, m^2-n^2 and m^2+n^2. The Platonic sequence itself can be derived by following the steps for 'splitting the square' described in Diophantus II.VIII. Neil Parker 21:06, 13 April 2007 (UTC). Edited 22:29, 14 April 2007

i dont know if this has been added yet but the diffrence between the lenth of the hypotnuse and a leg times the hypotnuse and leg added tohger is the shorter leg squared (c+b)(c-b)=a² which when foiled out becomes c²-b²=a². so when c-b=1 as in the case of a 3 4 5 b+c=a² —Preceding unsigned comment added by 66.229.198.17 (talk) 17:50, 2 March 2008 (UTC)

[edit] 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

[edit] God Created the Integers

"Given a primitive Pythagorean triple, find another". Beelzebub, October 22, 4004 BC

I am more and more convinced that God created the integers. After all, if He didn’t, who in Hell did? As evidence of His Hand at work, consider the following little game.

Choose 3 integers (a,b,c) and apply the following: Let p be twice their sum, and let ( |p-a|, |p-b|, |p+c| ) = (a’,b’,c’). For instance, if you start with (a,b,c) = (1,1,1) you find that (a’,b’,c’) = (5,5,7). “Ho hum” you were about to say, but change the rules slightly and you get the most spectacular result!!! Suppose the triple you start with is a primitive Pythagorean triple, a “PPT”. Also, change the rules slightly to allow one of a,b,c to be negative. There are four possibilities: (a,b,c) is (+++) or (-++) or (+-+) or (++-). Amazingly, each results in a new PPT !!!.

Draw a simple diagram of this result letting G represent a "given" PPT (a,b,c). To the left of G write the result obtained by using the (++-) case. To the right of G write the results obtained by using (-++), (+++) and (+-+). Connect the 4 triples to G with lines. We now have a tetrad of resultant triples produced as described above. For instance, if G is (3,4,5) we get (1,0,1) on the left, and (5,12,13), (21,20,29),and (15,8,17) on the right. We can identify (++-) as G's "parent", and the other 3 triples as G's "children".

Figure by L. Price, 2006

Clearly, this process can be infinitely extended by replacing G with one of the triples on the right. Doing so confirms the “parent–child” link, and produces 3 new triples to the right. A similar process moving leftward leads (in a finite number of steps) to (3,4,5). This wonderful result can be re-stated more formally as follows.

___________________________________________________________

Pythagorean Transformation Theorem

Let T: (a,b,c) --> (a’,b’,c’) be a transformation of positive integers as described in paragraph # 2 above. [ ie; Let p be twice their sum, and let ( |p-a|, |p-b|, |p+c| ) = (a’,b’,c’).]

-Then if (a,b,c) is a primitive Pythagorean triple, so is (a’,b’,c’).

-Transformation T is still valid when one of a,b,c is negative.

-T applied to a 'given' PPT, G, has four possible outcomes; ALL are PPT’s!!!.**

-Each PPT (except 3,4,5) has a single “parent” PPT, derived by applying T to the (++-) case, and 3 “children” derived from the remaining three cases (-++),(+++), (+-+).

-The “tree-like” structure of PPT’s shown in the figure can be infinitely extended by re-applying T to the triples on each end of the graph.

-Each PPT appears on the “Tree” precisely one time.

-Each PPT (except 3,4,5) has 2 “siblings”.

-Hypotenuses increase in magnitude with each generation. For a given tetrad (4 resultant triples), the largest hypotenuse belongs to the “middle child” (+++).

-All PPT’s descend in a finite number of generations from the same “Ancestor”, [ from PPT (3,4,5) ].

-The n-th generation on the Tree contains 3 ⁿ PPT’s.

** note: we exclude the "degenerate" triple (1,0,1). ________________________________________________________________

By means of this transformation we can provide all possible answers to Beelzebub's question, and construct the entire "Tree" given ANY PPT as our starting point. Moreover, we can specify how each PPT is related to any other. [What about Heronic triangles and other triangles with integer sides which are NOT right-angled? Except for the (++-) case, transformation T still applies!!!

Lee Price, Fairport, NY (tanutuva@rochester.rr.com) --Lee Price 17:42, 19 May 2006 (UTC)

Do the angles in triangles formed by pythagorean triples follow any pattern or series? If you go to very large integers, can a pythagorean triple be found to form any angle? If not, can one find one to be close to any angle (within a certain precision), or are whole ranges of angles impossible? TomViza 23:21, 18 January 2006 (UTC)

Since PPT's have rational sides, the trigonmetric functions are also rational, so measured in that way, we might find patterns of sorts, but not much has been done along these lines. The answer to your 2nd question is "no". Take a right triangle with sides equal to the square roots of 2, 3, and 5. If you check the half angle tangents, you'll see that they too are irrational (Obviously, no rational half-angle tangents q/p and q'/p' exist which are also irrational, so no PPT can have these angles.) You can, however, find a PPT which approximates these angles by finding four parameters q,q',p,p' which meet the following requirements; they share no common factors >1, one of q,p is even, p=q'+q , p'=q+p, AND the fractions q/p and q'/p' approximate to the desired degree of accuracy the half-angle tangents of your target triangle- a rather tall order! Your PPT would then have sides equal to: [2qp] , [q'p'], [qp'+q'p]. Lee Price--Lee Price 17:42, 19 May 2006 (UTC)

The whole "Pythagorean Triples as Integers" issue:

The page (incorrectly) says "The [Pythagorean theorem] states that any right triangle with integer side lengths yields a Pythagorean triple". I think that's where most of the confusion is coming from: Pythagorean triples are INTEGER solutions to the Pythagorean thoerem, not the ONLY solutions. Every other math site I've seen gets this correct: so I went ahead and corrected the entry. The definition of Pythagorean triples isn't the problem: it's the claim that all right triangles have integer sides. Andrew Gibson

Why is this limited to integers? What about Pythagorean triples such as 1:1:sqrt(2) or 1:sqrt(3):2?

Probably because Pythagoras and his followers didn't like irrational numbers. Vicki Rosenzweig

They didn't even have anything beyond the natural set of numbers. Even the concept of unity and zero were shrowded in mystery. The Pythagorean thoerem itself was 'likely' discoved using arrangements of small stones in L or square shaped patterns called "gnomons". Gregory S. Hayes

Okay, but nowadays no one has a problem with irrational numbers. Is this article about pythagorean triples in history or the concept of pythagorean triples in modern mathematics?

It's about the concept of Pythagorean triples, whose definition has never changed. These integral triples are much more interesting than the general concept of triples of real numbers (x,y,z) with x²+y²=z²; pretty much all you can say about the latter is that they form an infinite cone in 3-dimensional space. AxelBoldt

Correct. The remarks above about irrational numbers and 1:1:sqrt(2) and the like are naive. The equation x²+y²=z² is that of a conic section in the real projective plane in homogeneous coordinates, and individual points on that curve are not of interest unless it's because they are rational points.

I disagree. The triples for 45-45-90 and 30-60-90 seem important/unique enough to list. —Josh Lee 02:51, Apr 7, 2005 (UTC)

Guys, we're writing an encyclopedia here. Just find an authoratative reference that indicates pythagorean triples don't need to be integers, and we can include that info here. Otherwise it stays the way it is. --P3d0 03:42, July 23, 2005 (UTC)

I learned that there is a formula 2p, p2-1, p2+1 for generating Pythagorean triples but it is apparently weird. What is wrong with 3, 4, 5 and 5, 12, 13? Integer 5 takes the position of the lowest and the highest integer in these two examples. How can my formula explain this? 213.226.138.241 21:18, 25 Dec 2004 (UTC)

I presume you mean

2 p

p 2 - 1

, and

p 2 + 1

p = 2

gives the 3,4,5 sequence, while there is no value of

p

that gives 5,12,13 because the latter has no two elements that differ by 2. (Nobody ever claimed your formula gave all pythagorean triples, did they?) --P3d0 16:31, July 10, 2005 (UTC)

I learned from one of my school math teachers some other ways to generate pythagorean triples:

Given the integers n and x,
We have the expressions $2 x 2$ , $2 n x$ and $n 2$
The first number of the triple will be $2 x 2 + 2 n x$
The second one will be $2 n x + n 2$
The third one, $2 x 2 + 2 n x + n 2$

Example: Being $n = 3$ and $x = 5$ , we get:
1st number is $80$
2nd one is $39$
3rd one, $89$
Verifying: $392 + 80 2 = 89 2$

Another method: Given an integer n, the triple can be generated by the following two procedures:

First number (a) is $2 n + 1$
Second number (b) is $2 n 2 + 2 n)$
Third one is (c) $2 n 2 + 2 n + 1$
Example: if $n = 2$ we will obtain the triple 5, 12 and 13

Grim here, and I actually know what you're on about here, and i've edited it to be simpler. This one ^ is for triangles for which a^2+b^2=c^2, all sides' lengths are positive intigers, and a is odd. eg:

n	a	b	c
1	(2x1)+1=3	(2x1^2)+(2x1)=4	(2x1^2)+(2x1)+1=5
2	(2x2)+1=5	(2x2^2)+(2x2)=12	(2x2^2)+(2x2)+1=13
3	(2x3^2)+1=7	(2x3^2)+(2x3)=24	(2x3^2)+(2x3)+1=25

as you can see, c is always b+1 in this 'family' of triples, and b+c=a^2, if i'm not mistaken. there are terms of n for the perimter & area, but i cannot put them down fow the moment, as i'm supposed to be doing this for maths coursework... i hate coursework. sorry about editing in the middle of your idea, but it was easier than copying the whole thing out again. -Grim- 06:57, 20 March 2007 (UTC)

First number is $4(n + 1)$
Second one is $4(n + 1) 2 - 1$
Third one is $4(n + 1) 2 + 1$
Example: if $n = 5$ , we generate the triple 24, 143 and 145

I apologize for my poor editing --Marlino 05:06, 21 July 2005 (UTC)

This article started to look like an incoherent collection of random factoids, so I decided to clean it up. I moved the material here in case anyone wants to take care of these ugly ducklings.

Deleted this trivial observation (too trivial to waste any markup on, apparently):

if m1*n1=m2*n2, then the calculated triples will have an equal number.

and a "good idea" that remains unused:

A good starting point for exploring Pythagorean triples is to recast the original equation in the form:

b² = (c − a)(c + a)

and a random observation. Many similar patterns exist but aren't mentioned either:

It is interesting to note that there are more than one primitive Pythagorean triple with the same lowest integer, the first example is for 20, which is the lowest integer of two primitive triples: 20 21 29 and 20 99 101.

and a further example designed to impress, rather than to explain:

By contrast the number 1229779565176982820 is the lowest integer in exactly 15386 primitive triples, the smallest and largest triples it is part of are:

1229779565176982820

1230126649417435981

1739416382736996181

and

1229779565176982820

378089444731722233953867379643788099

378089444731722233953867379643788101.

For the curious, consider the prime factorisation

1229779565176982820 = 2² × 3 × 5 × 7 × 11 × 13 × 17 × 19 × 23 × 29 × 31 × 37 × 41 × 43 × 47.

The number of prime factors is related to the large number of primitive Pythagorean triples. Note that there are larger integers that are the lowest integer in an even greater number of primitive Pythagorean triples.

and a statement that's true but off-topic:

Fermat's last theorem states that non-trivial triples analogous to Pythagorean triples but with exponents higher than 2 don't exist.

—Herbee 22:59, 4 September 2005 (UTC)

The smallest Pythagorean triple in which the hypotenuse and the sum of the arms are squares

(e.g. a+b = a square integer and c= a square integer is

a = 4565486027761 b = 1061652293520 c = 4687298610289

proof:

the sum of a+ b = 5,627,138,321,281

the sqrt(a+b)= 2,372,159

the sqrt (c) = 2,165,017

Pythagorean Triangles, by Waclaw Sierpinski, Dover Publications, Inc., 2003, ISBN 0-486-43278-5

Triples can also be parametrized using this less well-known approach: 2xy = $z 2$ , x,y,z > 0 where the following relations hold

x = c-b, y = c-a, z = a+b-c and a = x+z, b = y+z, c = x+y+z

Pythagorean triples can then be generated by choosing any even integer z.

x and y are any two factors of $z 2 / 2$

Example choose z = 6 then

z 2 / 2

=18 = z

the threefactor pairs of 18 are: (18,1) , (2,9), and (6,3) . All three factor pairs will produce triples using the above equations.

z=6 x=18 y=1 produces the triple a=18+6=24 b=1+6=7 c=18+1+6=25

z=6 x=2 y=9 produces the triple a= 2+6=8 b=9+6=15 c= 2+9+6=17

z=6 x=6 y=3 produces the triple a= 6+6=12 b=3+6=9 c= 6+3+6=15

[edit] Pythagorean triple

In reference to your webpage Pythagorean Triple in particular primitve triples, m and n can both be odd if you take the formulas from below(on the same webpage Pythagorean Triple) a=m,b=(m^2-1)/2,c=(m^2+1)/2 and substitute n for 1 to get a=m*n,b=(m^2-n^2)/2,c=(m^2+n^2)/2 where m and n are both odd and relatively prime m being the greater. This contribution is from Joe Mc.207.200.116.12 19:32, 14 October 2006 (UTC)

[edit] mis-wording?

Ok, I've been scratching my head. What do these three lines mean?

For each natural number n, there exist n Pythagorean triples with different hypotenuses and the same area.
For each natural number n, there exist at least n different Pythagorean triples with the same leg a, where a is some natural number
For each natural number n, there exist at least n different triangles with the same hypotenuse.

Tim

Comment: See the book: "Pythagorean Triangles" by Waclaw Sierpinski (Dover Publications).

Direct quotes from the book:

"It is easy to prove that for each natural number n there exit at least n different pythagorean triangles with the same arm a, where a is some natural number." (page 30)

"For each natural number n there exist at least n different triangles with the same hypotenuse." (page 31)

"For each natural number n there exist n pythagorean triangles with different hypotenuses and the same area." (page 37)

the book explains the math in more detail.

Example,

If n=50, then there are at least 50 different triples with the same arm, a If n=50 then there are at least 50 different triples with the same hypotenuse if n=50 then there at at least 50 different triples with different hypotenuses and the same area etc. for each integer

So is there a difference between a Pythagorean triple, which is a set of three integers, and an integal Pythagorean triangle, which has two legs and an hypotenuse? Should the terminology be fixed in this article? --MathMan64 17:36, 1 November 2006 (UTC)

Response: A pythagorean triple is not just a set set of three integers. The term "pythagorean triple" means a set of three integers which form a right triangle which by definition has two legs and a hypotenuse. Hence, the term is correctly used so there is no need to fix the terminology in this article. See definitions of pythagorean triples at the following:

http://www.math.rutgers.edu/~erowland/pythagoreantriples.html http://mathworld.wolfram.com/PythagoreanTriple.html http://zimmer.csufresno.edu/~larryc/proofs/proofs.construct.html http://homepage.smc.edu/kennedy_john/PYTHAGT.PDF http://library.thinkquest.org/2647/geometry/glossary.htm http://math.temple.edu/~wds/homepage/diophant.pdf http://math.arizona.edu/~nfrogers/talks/pythagoras.pdf http://www.math.ou.edu/~dmccullough/teaching/pythagoras2.pdf --Bronson Gardner

[edit] Prime numbers??

I wanted to know if the following has been disproven:

In any leg-hypotenuse pythagorean triple, at least one of a, c, a+b, and b-a is prime.

To check, we have:

3, 4, 5
5, 12, 13
7, 24, 25
9, 40, 41
11, 60, 61
13, 84, 85
15, 112, 113
17, 144, 145
19, 180, 181

If we restrict this to a and c, the next triple disproves this variant:

21, 220, 221 (note that neither 21 = 3*7 nor 221 = 13*17 is prime)

However, if we expand the problem to include a+b, 21+220=241, which is prime.

23, 264, 265
25, 312, 313

The next triple after this disproves the variant that includes a, c, and a+b but not b-a

27, 364, 365 (27+364=391 = 17*23)

However, expanding this to include a-b gives 364-27=337 which is prime.

29, 420, 421
31, 480, 481
33, 544, 545 (33+544=577)
35, 612, 613
37, 684, 685
39, 760, 761
41, 840, 841
43, 924, 925
45, 1012, 1013
47, 1104, 1105
49, 1200, 1201
51, 1300, 1301
53, 1404, 1405
55, 1512, 1513 (55+1512=1567)
57, 1624, 1625 (57+1624=1681, = 41*41 but 1624-57=1567)
59, 1740, 1741
61, 1860, 1861
63, 1984, 1985 (63+1984=2047, = 23*89 and 1984-63=1921 = 17*113) and this disproves this variant (specifically the one including a, c, a+b, and b-a.) Any variant for which this is an open problem?? Georgia guy 22:02, 16 April 2007 (UTC)

[edit] A Pythagorean Triple which approximates 30,60,90

This follows from a contribution in the topic entitled 'God Created the integers'. Somebody asked if you could approximate any right triangle with a Pythagorean triple - well I tried approximating a 60 degree right triangle and came up with the following:

6273,10864,12545

This was found by simply drawing a 30 degree line on graph paper and looking to find the places where the line intersected the corners of 1mm squares. Hence you find candidate m,n pairs for the formula:

2mn, m^2-n^2, m^2+n^2. [with n/m ≈ tan(60/2) = tan(30)]

The pair which generates the above triple is 97,56 and this is the best I could find using the above method. But you can easily use a spreadsheet to find even better approximations. It is quite fun for a school project on PTs since the same technique can be used for other angles.

Neil Parker 18:43, 4 June 2007 (UTC)

Here are the first twenty Pythagorean triples (a_k, b_k, h_k) such that the hypotenuse and twice the short leg differ by one. Note that all such triples are primitive. Your example is the seventh in the sequence:

(3, 4, 5)

(8, 15, 17)

(33, 56, 65)

(120, 209, 241)

(451, 780, 901)

(1680, 2911, 3361)

(6273, 10864, 12545)

(23408, 40545, 46817)

(87363, 151316, 174725)

(326040, 564719, 652081)

(1216801, 2107560, 2433601)

(4541160, 7865521, 9082321)

(16947843, 29354524, 33895685)

(63250208, 109552575, 126500417)

(236052993, 408855776, 472105985)

(880961760, 1525870529, 1761923521)

(3287794051, 5694626340, 6575588101)

(12270214440, 21252634831, 24540428881)

(45793063713, 79315912984, 91586127425)

(170902040408, 296011017105, 341804080817)

The short leg is given by:

$a_k=\frac{s_k - (-1)^k}{3}$

where

$s_k=\frac{(2+\sqrt 3)^{k+1}+(2-\sqrt 3)^{k+1}}{2} - (-1)^k$ .

The the long leg b_k is given by:

$b_k = \frac{(2+\sqrt 3)^{k+1} - (2-\sqrt 3)^{k+1}}{2\sqrt 3}$ .

Finally, the hypotenuse h_k is given by:

$h_k = \frac{2s_k+(-1)^k}{3} = 2a_k +(-1)^k$ .

Evidently, these triples are successively better approximations to a 30-60-90 triangle. -- Chuck 04:49, 5 June 2007 (UTC)

Some solutions of this type can be found by solving a Pell equation m² = 3 n² ± 1, with short side m² - n² and hypotenuse is m² - n². —David Eppstein 09:23, 5 June 2007 (UTC)

You must mean for the hypotenuse to be m² + n². To be more precise, for odd k = 2j - 1, the short leg a_k = m_j² - n_j² and hypotenuse h_k = m_j² + n_j² where (m_j,n_j) is the j-th solution of the Pell equation m² = 3 n² + 1. The first ten of these solutions (m_j, nm_j) are:

(2, 1)

(7, 4)

(26, 15)

(97, 56)

(362, 209)

(1351, 780)

(5042, 2911)

(18817, 10864)

(70226, 40545)

(262087, 151316)

Note that n_j = b_j-1, the long leg of the (j-1)^st triple in the sequence given previously.

For even k = 2 j, the long leg b_k = m_j² - n_j² and hypotenuse h_k = m_j² + n_j² where m_j = b_j and n_j = 'b_j-1. in this case, the the numbers (m_j - 2 n_j) and n_j satisfy the Pell equation (m_j - 2 n_j)² = 3 n_j² + 1. The first ten of these solutions (m_j, n_j) are:

(4, 1)

(15, 4)

(56, 15)

(209, 56)

(780, 209)

(2911, 780)

(10864, 2911)

(40545, 10864)

(151316, 40545)

(564719, 151316)

The key to obtaining these explicit solutions is the observation that the numbers n_j = b_j-1 (in both cases) satisfy the two-term recurrence relation:

$n_{j+1} = 4n_j - n_{j-1}\,$ .

-- Chuck 15:12, 5 June 2007 (UTC). Addendum and minor corrections -- Chuck 13:57, 6 June 2007 (UTC)

[edit] Pythagoras' Theorem and the Platonic Sequence: A Common Genesis

Pythagoras' Theorem

Figure 1

In the diagram COD is a diameter with O the centre. Right triangle OAB has sides x,y,r as shown.Let angle AOB = θ and , since this is an external angle of isosceles triangle DOB, angle D is θ/2.

Angle CBD stands on the diameter and is therefore $900 .$

Angle C is complementary to angle D and to angle ABC making the latter also equal to θ/2.

Triangles ACB and ABD are right triangles with angle D = angle ABC = θ/2. Therefore they are similar. Hence:

$\begin{align}\frac{AB}{AC}&=\frac{AD}{AB}\\ \therefore \frac{y}{r-x}&=\frac{r+x}{y}\\ \Rightarrow y^2&=(r-x)(r+x)=r^2-x^2\\ \Rightarrow r^2&=x^2+y^2 \end{align}$

The Platonic Sequence.

Figure 2

In the same diagram (Figure 2) let AC = 1 unit. (compared with Figure 1 we are simply scaling everything by a factor r – x). Then:

$\begin{align} AB&=t\ where\ t=\cot(\tfrac{\theta}{2})\quad [\tfrac{\theta}{2}\in(0,45^\circ]\Rightarrow t\geqslant 1]\\ AD&=AB. \cot(\tfrac{\theta}{2})=t^2\\ CD&=CA+AD=t^2+1\\ OB&=OD=\tfrac{CD}{2}=\tfrac{t^2+1}{2}\\ OA&=OC-1=\tfrac{t^2-1}{2} \end{align}$

It can be seen that the scaled right triangle OAB in Figure 2 has sides which fit the Platonic sequence and since Figure 1 and Figure 2 are otherwise identical, it should be clear that the Platonic sequence is in fact a core expression of the Pythagorean identity. In particular rational triples arise if t is a rational number m/n:

$\begin{align} AB &= t = m/n\\ OA &= \frac{t^2-1}{2}=\cfrac{\tfrac{m^2}{n^2}-1}{2}=\frac{m^2-n^2}{2n^2}\\ OB &= \frac{t^2+1}{2}=\cfrac{\tfrac{m^2}{n^2}+1}{2}=\frac{m^2+n^2}{2n^2}\\ \end{align}$

To obtain Pythagorean triples with integer sides we simply scale the above by a factor $2 n 2$ to give:

$\begin{align} AB&=2mn\\ OA&=m^2-n^2\\ OB&=m^2+n^2 \end{align}$

All Pythagorean triples are therefore scaled versions of similar (ie equi-angular) Platonic sequence triangles which have rational sides. If by definition Platonic sequence triangles must have integer sides, we should rather use the term half angle cot sequence in the above context.

Figure 3

As one would expect there is a corresponding half angle tan sequence: refer to figure 3 and let AD = 1 unit. (compared with Figure 1 we are now scaling everything by a factor r + x). Then:

$\begin{align} AB&=t\ where\ t=\tan(\tfrac{\theta}{2})\quad [\tfrac{\theta}{2}\in[0,45^\circ]\Rightarrow 0\leqslant t \leqslant1]\\ AC&=AB. \tan(\tfrac{\theta}{2})=t^2\\ CD&=DA+AC=1+t^2\\ OB&=OD=OC=\tfrac{CD}{2}=\tfrac{1+t^2}{2}\\ OA&=OC-t^2=\tfrac{1-t^2}{2} \end{align}$

Once again rational triples arise if t is a rational number m/n where m<n in this instance.

Neil Parker 16:35, 25 June 2007 (UTC).

Edited to include the half angle tan sequence. Please note that I also made corrections in light of the query below. Neil Parker 10:48, 9 July 2007 (UTC)

What do yo mean by the last sentence? What are "derivatives" of the Platonic sequence? Are you trying to imply that any Pythagorean triple has the form (a,b,c), where c = m² + n² with b and c being m² - n² and 2mn in some order? The latter is true of all primitive Pythagorean triples and some but not all other Pythagorean triples. For example (6,8,10) = ( 2mn, m²-n², m²+n²), where m = 3 and n = 1, but (9,12,15) cannot occur in this fashion for any choice of natural numbers m, n (because 15 cannot be expressed as m²+n² at all). -- Chuck 18:34, 25 June 2007 (UTC)

Hope the above ammendment clarifies things! Neil Parker 09:07, 26 June 2007 (UTC)

[edit] "Below is a formula created by Matthew Irvine"

The following culmination of a sequence of edits by 124.180.159.248 was removed to here.

===Below is a formula created by Matthew Irvine===

This formula returns a Pythagorean triple

$a=de , b={d(e^2-1)\over 2} , c=b+d$

Simply choose any natural number greater that two for e. Then choose any natural number for d if e is odd or any even number for d if e is even.

For the primitive triples simply use the lowest possible d (If e is even that will be two and if e is odd that will be one).

For example if e=3 then d=1 meaning a=3, b=4, c=5 or if e=4 then d=2 meaning a=8, b=15, c=17

It is really just a rephrasing of the material in the section "A special case: the Platonic sequence" in the main article. -- Chuck 20:33, 9 July 2007 (UTC)

[edit] Diophantus II.VIII and the Platonic Sequence

The purpose of the following article is to demonstrate how the Platonic sequence can be derived by following the steps given in Diophantus II.VIII.

Diophantus II.VIII: Intersection of line CB and the circle has rational solutions for x and y

"To divide a square into a sum of two squares
To divide 16 into a sum of two squares

Let the first summand be $x 2$ and thus the second $16 - x 2$ . The latter is to be a square. I form the square of the difference of an arbitrary multiple of x diminished by the root of 16, that is, diminished by 4. I form, for example, the square of 2x-4. It is $4 x 2 - 16 x + 16$ . I put this expression equal to $16 - x 2$ and subtract 16. In this way I obtain $5 x 2 = 16 x$ , hence $x = 16 / 5$ .

Thus one number is 256/25 and the other 144/25. The sum of these numbers is 16 and each summand is a square."

Without in any way detracting from the method described above, we generalize to solve the problem for any given square which we will represent algebraically as $a 2$ . Also since Diophantus refers to an "arbitrary multiple of x" we will use the letter t to represent that coefficient. Then:

$\begin{array}{lcl} \qquad (tx-a)^2 & = & a^2-x^2\\ \Rightarrow t^2x^2-2atx+a^2 & = & a^2-x^2\\ \Rightarrow x^2(t^2+1)& = & 2atx\\ \Rightarrow x & = & \frac{2at}{t^2+1}\quad or \quad x=0\\ \end{array}$

"Thus one number is $\left(\tfrac{2at}{t^2+1}\right)^2$ and the other is $\left(\tfrac{a(t^2-1)}{t^2+1}\right)^2$ . The sum of these numbers is $a 2$ and each summand is a square."

Diophantus II.VIII: Generalized solution in which triangle OAB is a rational triple if line CB has a rational gradient t

The specific results obtained by Diophantus may be obtained by substituting a=4 and t=2 in the above algebraic expressions. In effect his solution yields the following rational triple representing the sides of triangle OAB in the diagram (it is assumed a and t are both rationals):

$\left[ a; \frac{2at}{t^2+1}; \frac{a(t^2-1)}{t^2+1}\right]=\left[ \frac{20}{5};\frac{16}{5};\frac{12}{5} \right]=\frac{4}{5} \left[5;4;3\right]$

We see that Diophantus' particular solution is in fact a subtly disguised 3,4,5 triple but there is no doubt he could obtain any of an infinity of triples simply by changing the value of that "arbitrary multiple of x".

The algebraic solution needs only one additional algebraic step to arrive at the Platonic sequence $[\tfrac{t^2+1}{2};t;\tfrac{t^2-1}{2}]$ and that is to multiply all sides of the above triple by a factor $\quad \tfrac{t^2+1}{2a}$ . Notice also that if a=1, (unit circle) the sides (OB,OA,AB) reduce to:

$\left[ 1; \frac{2t}{t^2+1}; \frac{t^2-1}{t^2+1}\right]$

In modern notation this is just $(1, s i n θ, c o s θ)$ written in terms of the half angle cot which in the particular example given by Diophantus has a value of 2, the "arbitrary coefficient of x". Intriguing that this "arbitrary coefficient" has become the cornerstone of the Diophantine triple generator!

References:

Diophantus of Alexandria. Arithmeticorum Lib II Quaestio VIII

Heath, Sir Thomas L. Diophantus of Alexandria: A Study in the History of Greek Algebra. Dover.
Bashmakova, Isabella Grigoryevna. Diophantus and Diophantine Equations . The Mathematical Association of America. USA, 1997.
Both as quoted by Aaron Zerhusen, Chris Rakes, & Shasta Meece.Diophantine Equations

Further Reading:
Fermat's Last Theorem and Diophantus II.VIII

Neil Parker 15:35, 13 July 2007 (UTC)

The page states properties of primitive Pythagorean triples:

There exist infinitely many Pythagorean triples in which the hypotenuse and the longer of the two legs differ by exactly one.
There exist infinitely many Pythagorean triples in which the hypotenuse and the longer of the two legs differ by exactly two.

- These two statements are also true for PRIMITIVE Pythagorean triples!

There exist infinitely many Pythagorean triples in which the hypotenuse and the longer of the two legs differ by exactly three.

- But there are no PRIMITIVE Pythagorean triples in which the hypotenuse and a leg differ by exactly a prime number greater than 2.
AdrianSavage (talk) 19:43, 25 November 2007 (UTC)

I'll add that last point you make to the page. It's vastly more interesting than "There are infinitely many -- albeit non-primitive -- triples in which b and c differ by 3", which is just obvious, since once you know there are infinitely many primitive triples in which b and c differ by 1, you can multiply every term of the triple by 3. 91.107.153.28 (talk) 23:48, 21 February 2008 (UTC)

[edit] 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.

[edit] Suggested links

Maybe you will agree on adding these links:

"Pythagorean Triples Star" by Enrique Zeleny, The Wolfram Demonstrations Project.
"Plotting Pythagorean Triples" by Enrique Zeleny, The Wolfram Demonstrations Project. —Preceding unsigned comment added by 200.92.215.174 (talk) 00:18, 30 March 2008 (UTC)

69.252.161.255 (talk) 05:29, 21 March 2008 (UTC)