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| Besides Euclid's formula, many other '''formulas for generating [[Pythagorean triple]]s''' have been developed. these formulas are used variously to generate pythagorean triples.
| | There were surveys produced and discovered out that about 80% of females are dissatisfied with their breasts sizes. So numerous females wish that they have firmer and sexier breasts.<br><br>A single of the known possibilities is to undergo breasts surgery. Surgery is one of the approaches that can give ladies bigger and firmer breasts. This approach is costly. [http://www.revish.com/people/weightlossoha/ Close Window] contains more about why to consider this thing. This dazzling [https://storify.com/openbrowsera981/wonderful-getting-older-recommendation-that-can-wo homepage] article has collected influential cautions for where to look at it. And there are a lot of dangers accompanied with surgery which drives lots of ladies to wish there are natural way to boost their breasts.<br><br>With the eagerness of most girls to have firmer and sexier breasts, they are turning to organic breast enhancement pill that can give them larger breasts. Organic breast enhancement pill can give females bigger and firmer breast with out undergoing the risky and pricey surgery. So they can get larger and firmer breasts without having the feeling of the risky surgery.<br><br>These pills are made up of herbs, so they are all-natural merchandise that can give ladies bigger breasts. Learn further on this related article directory - Click here: [http://about.me/openbrowseryacht image]. But of course, you nevertheless have to be careful, given that not these whole natural breast enhancement pills can work, so you want to find the a single that can genuinely work for you.<br><br>It is beneficial to make a research. Know the product 1st prior to purchasing and using it. You need to have to get the appropriate data about the product, such as how it was produced, its substances, exactly where it came from, how it functions, how effective it is and so on.<br><br>If you are looking for natural breast enhancement pill online, there are masses of sites that offer all-natural breast enhancement pills, so you require to discover a reliable internet site in order to make confident to have high-quality product. A excellent internet site can give their customers with useful info that are valuable to consumers, given that they require to know the item and how it operates.<br><br>Producing a research can be beneficial to you because there are firms out there that creates products that do not function at all, which would be a waste of money and it can even harm your wellness. So it is better to be alert and allocate small time in finding the proper product that can truly give you firmer and fuller breasts.<br><br>Of course, the primary purpose why you are searching for the proper organic breast enhancement pill is to have fuller and firmer breast so you can feel sexy and feel wonderful about yourself. You want to properly get firmer and fuller breasts with out any adverse reactions, so creating easy study is an suitable way to uncover it.<br><br>Organic breast enhancement pill is a great way to boost your breast, given that you do not have to face the risks of having unfavorable side effects that surgery can give. But in order to get the proper a single, you want to make easy research.<br><br>As you buy a all-natural breast enhancement pill, you want to make certain that you will adhere to meticulously the directions, so read the label and understand it. In undertaking so, you can see exceptional benefits..<br><br>When you have virtually any concerns about exactly where and the way to utilize high deductible health plan ([http://gulliblebottlen24.pen.io please click the next website page]), you can call us from the web page. |
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| ==Euclid's, Pythagoras', and Plato's formulas==
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| Euclid's, Pythagoras' and Plato's formulas for calculating triples have been described here: {{main|Pythagorean triple}}
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| The methods below appear in various sources, often without attribution as to their origin.
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| ==Generating triples when one side is known==
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| The following method is a direct algebraic manipulation of the Euclid equations.
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| Start with any even integer <math>b</math>, and use the relation <math>b = 2mn</math> from the Euclid formula. Identify all factor-pairs (m,n) of <math>b</math> and use the Euclid equations to calculate the remaining sides of the triple.
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| Examples:
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| Let <math>b=24 </math> (e.g. the known side is even)
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| <math>24 = 2mn </math> so that <math>12 = mn </math> . The factor pairs (m,n) of 12 are (12,1), (6,2) and (4,3). The three possible triples are therefore:
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| : <math> a = (m^2-n^2) \,,\ b =2mn \,,\ c = (m^2 + n^2)</math>
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| : <math> a=12^2-1^2 =143 \,,\ b = 24 \,,\ c = (12^2 + 1^2)=145</math>
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| : <math> a= 6^2-2^2 =32 \,,\ b = 24 \,,\ c = (6^2 + 2^2)=40</math>
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| : <math> a = 4^2-3^2 =7 \,,\ b = 24 \,,\ c = (4^2 + 3^2)=25</math>
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| Let <math>b =35</math> (e.g. the known side is odd)
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| The two unknown sides could also be calculated by making use of the relation <math> a = (m^2-n^2)</math> . This would be a factoring exercise in finding the difference of two squares, but a simpler approach is to multiply the known side by two and continue as before:
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| : <math>70 = 2mn </math> so that <math>35 = mn. </math>
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| The factor pairs (''m'',''n'') of 35 are (35,1), (7,5).
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| The two triples are therefore (note that is necessary to remove the factor of 2 which was introduced):
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| : <math> a=(35^2-1^2)/2=612 \,,\ b = 70/2=35 \,,\ c = (35^2 + 1^2)/2=613 </math>
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| : <math> a=(7^2-5^2)/2=12 \,,\ b = 70/2=35 \,,\ c = (7^2 + 5^2)/2=37 </math>
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| ==Fibonacci's method==
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| [[Leonardo of Pisa]] (c. 1170 – c. 1250) described this method<ref>Fibonacci, Leonardo Pisano, (1225), [[The Book of Squares|Liber Quadratorum]].</ref><ref>Fibonacci, Leonardo Pisano . The Book of Squares (Liber Quadratorum). An annotated translation into modern English by L. E. Sigler. (1987) Orlando, FL: Academic Press. [http://www.elsevierdirect.com/ISBN/9780126431308/Leonardo-Pisano-Fibonacci ISBN 978-0-12-643130-8]</ref> for generating primitive triples using the sequence of consecutive odd integers
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| <math>1,3,5,7,9,11,\ldots</math>, and the fact that the sum of the first <math>n</math> terms of this sequence is <math>n^2</math>. If <math>k</math> is the <math>n</math>-th member of this sequence then <math>n=(k+1)/2</math>.
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| Choose any odd square number <math>k</math> from this sequence (<math>k=a^2</math> ) and let this square be the <math>n</math>-th term of the sequence. Also, let <math>b^2</math> be the sum of the previous <math>n-1</math> terms, and let <math>c^2</math> be the sum of all <math>n</math> terms . Then we have established that <math>a^2+b^2=c^2</math> and we have generated the primitive triple [ ''a, b, c'']. This method produces an infinite number of primitive triples, but not all of them.
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| EXAMPLE:
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| Choose <math>k=9=3^2=a^2</math> . This odd square number is the fifth term of the sequence, because <math>5=n=(a^2+1)/2</math>. The sum of the previous 4 terms is <math>b^2 = 4^2</math> and the sum of all <math>n=5</math> terms is <math>c^2 = 5^2</math> giving us <math>a^2+b^2=c^2</math> and the primitive triple [ ''a, b, c''] = [ 3, 4, 5].
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| ==Progressions of whole and fractional numbers==
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| The German monk and mathematician [[Michael Stifel]] published the following method in 1544.<ref>Stifel, Michael, (1544),
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| [http://mathdl.maa.org/mathDL/46/?pa=content&sa=viewDocument&nodeId=2591&bodyId=3752 Arithmetica Integra].</ref><ref>{{Cite journal |author=Ozanam, Jacques | url=http://books.google.com/?id=s_IJAAAAMAAJ&pg=RA3-PA49&lpg=RA3-PA49&dq=%22progression+of+whole+and+fractional+numbers%22&q=%22progression%20of%20whole%20and%20fractional%20numbers%22 |title=Recreations in Mathematics and Natural Philosophy |publisher=G. Kearsley|volume=1|page=49 |year=1814|accessdate=2009-11-19 |postscript=. }}</ref>
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| Consider the progression of whole and fractional numbers:
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| <math>1\tfrac{1}{3},\text{ }2\tfrac{2}{5},\text{ }3\tfrac{3}{7},\text{ }4\tfrac{4}{9},\ldots </math>
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| The properties of this progression are:
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| (a) the whole numbers are those of the common series and have unity as their common difference; (b) the numerators of the fractions, annexed to the whole numbers, are also the natural numbers; (c) the denominators of the fractions are the odd numbers, <math>3,\text{ }5,\text{ }7,\text{ }9,</math> etc.
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| To calculate a Pythagorean triple select any term of this progression and reduce it to an improper fraction. For example, take the term <math>3\tfrac{3}{7}</math>. The improper fraction is <math>\tfrac{24}{7}</math>. The numbers 7 and 24 are the sides, ''a'' and ''b'', of a right triangle, and the hypotenuse is one greater than the largest side. For example:
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| : <math>1\tfrac{1}{3}\text{ }\xrightarrow{\text{yields}}\text{ }[3,4,5],\text{ 2}\tfrac{2}{5}\text{ }\xrightarrow{\text{yields}}\text{ }[5,12,13],\text{ 3}\tfrac{3}{7}\text{ }\xrightarrow{\text{yields}}\text{ }[7,24,25],\text{ 4}\tfrac{4}{9}\text{ }\xrightarrow{\text{yields}}\text{ }[9,40,41],\text{ }\ldots</math>
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| [[Jacques Ozanam]]<ref>Ozanam, Jacques, (1844). Science and Natural Philosophy: Dr. Hutton’s Translation of Montucla’s edition of Ozanam, revised by Edward Riddle, Thomas Tegg, London. [http://ebooks.library.cornell.edu/cgi/t/text/pageviewer-idx?c=math;cc=math;q1=ozanam;rgn=full%20text;idno=07160001;didno=07160001;view=image;seq=7;page=root;size=50 Read online- Cornell University]</ref> republished Stifel’s sequence in 1694 and added the similar sequence <math>1\tfrac{7}{8},\text{ }2\tfrac{11}{12},\text{ }3 \tfrac{15}{16},\text{ }4\tfrac{19}{20},\ldots</math> with terms derived from <math>n+\tfrac{4n+3}{4n+4}</math>. As before, to produce a triple from this sequence, select any term and reduce it to an improper fraction. The numerator and denominator are the sides, ''a'' and ''b'', of a right triangle. In this case, the hypotenuse of the triple(s) produced is 2 greater than the larger side. For example:
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| : <math>1\tfrac{7}{8}\xrightarrow{\text{yields}}[15,8,17],2\tfrac{11}{12}\xrightarrow{\text{yields}}[35,12,37],3\tfrac{15}{16}\xrightarrow{\text{yields}}[63,16,65],4\tfrac{19}{20}\xrightarrow{\text{yields}}[99,20,101],\ldots </math>
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| Together, the Stifel and Ozanam sequences produce all primitive triples of the '''''Plato''''' and '''''Pythagoras''''' families respectively. The '''''Fermat''''' family must be found by other means.
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| <math>Plato:c-b=1,\quad \quad Pythagoras:c-a=2,\quad \quad Fermat:\left| a-b \right|=1</math> | |
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| ==Dickson's method==
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| [[L. E. Dickson|Leonard Eugene Dickson]] (1920)<ref>Dickson, L. E. (1920), ''[[History of the Theory of Numbers]], Vol.II. Diophantine Analysis'', Carnegie Institution of Washington, Publication No. 256, 12+803pp [http://www.archive.org/stream/historyoftheoryo02dickuoft#page/168/mode/2up Read online - University of Toronto]</ref> attributes to himself the following method for generating Pythagorean triples. To find integer solutions to <math>x^2+y^2=z^2</math>, find positive integers ''r'', ''s'', and ''t'' such that <math>r^2=2st</math> is a square.
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| Then:
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| : <math> x = r + s \,,\,y = r + t \,,\, z = r + s + t. </math>
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| From this we see that <math>r</math> is any even integer and that ''s'' and ''t'' are factors of <math>\tfrac{r^2}{2}</math>. All Pythagorean triples may be found by this method. When ''s'' and ''t'' are coprime the triple will be primitive. A simple proof of Dickson's method has been presented by Josef Rukavicka (2013).<ref>Rukavicka, J. (2013), ''Dickson's Method for Generating Pythagorean Triples Revisited'', European Journal of Pure and Applied Mathematics ISSN 1307-5543, Vol. 6, No. 3 (2013) p.363-364, [http://ejpam.com/index.php/ejpam/article/view/1844/309 online]</ref>
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| Example: Choose ''r'' = 6. Then <math>\tfrac{r^2}{2} = 18</math>.
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| The three factor-pairs of 18 are: (1, 18), (2, 9), and (3, 6). All three factor pairs will produce triples using the above equations.
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| :''s'' = 1, ''t'' = 18 produces the triple [7, 24, 25] because ''x'' = 6 + 1 = 7, ''y'' = 6 + 18 = 24, ''z'' = 6 + 1 + 18 = 25.
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| :''s'' = 2, ''t'' = 9 produces the triple [8, 15, 17] because ''x'' = 6 + 2 = 8, ''y'' = 6 + 9 = 15, ''z'' = 6 + 2 + 9 = 17.
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| :''s'' = 3, ''t'' = 6 produces the triple [9, 12, 15] because ''x'' = 6 + 3 = 9, ''y'' = 6 + 6 = 12, ''z'' = 6 + 3 + 6 = 15. (Since ''s'' and ''t'' are not coprime, this triple is not primitive.)
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| ==Generalized Fibonacci sequence==
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| ===I.===
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| For Fibonacci numbers starting with F<sub>1</sub>=0 and F<sub>2</sub>=1 and with each succeeding Fibonacci number being the sum of the preceding two, one can generate a sequence of Pythagorean triples starting from (''a''<sub>3</sub>, ''b''<sub>3</sub>, ''c''<sub>3</sub>) = (4, 3, 5) via
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| :<math>(a_n, b_n, c_n) = (a_{n-1}+b_{n-1}+c_{n-1}, \, F_{2n-1}-b_{n-1}, \, F_{2n})</math>
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| for ''n'' ≥ 4. See also [[Special right triangles#Fibonacci triangles|Fibonacci triangles]]. | |
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| ===II.===
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| A Pythagorean triple can be generated using any two positive integers by the following procedures using generalized [[Fibonacci sequence]]s.
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| For initial positive integers ''h''<sub>''n''</sub> and''h''<sub>''n''+1</sub>, if ''h''<sub>''n''</sub>+''h''<sub>''n''+1</sub>=''h''<sub>''n''+2</sub> and ''h''<sub>''n''+1</sub>+''h''<sub>''n''+2</sub>=''h''<sub>''n''+3</sub>, then
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| :<math>(2h_{n+1} h_{n+2}, h_nh_{n+3}, 2h_{n+1}h_{n+2}+h_n ^2)</math>
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| is a Pythagorean triple.<ref>Horadam, A. F., "Fibonacci number triples", ''[[American Mathematical Monthly]]'' 68, 1961, 751-753.</ref> | |
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| ===III.===
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| The following is a [[Matrix (mathematics)|matrix]]-based approach to generating primitive triples with generalized Fibonacci sequences.<ref name=Bernhart>{{cite arxiv|last=Bernhart, Frank R. and Price, H. Lee|title=Heron's formula, Descartes circles, and Pythagorean triangles|year=2005|pages=29|eprint=math/0701624v1|class=math.MG }}</ref> Start with a 2 × 2 array and insert two coprime positive integers '''''( q,q' )''''' in the top row. Place the even integer (if any) in the {{nowrap|left-hand}} column.
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| : <math>
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| \left[ {\begin{array}{*{20}c}
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| q & {q'} \\
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| \bullet & \bullet
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| \end{array}} \right]
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| </math>
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| Now apply the following "Fibonacci rule" to get the entries in the bottom
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| row:
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| : <math>
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| \begin{array}{*{20}c}
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| q' + q = p \\
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| q + p = p'
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| \end{array} \to \left[ {\begin{array}{*{20}c}
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| q & q' \\
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| p & p'
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| \end{array}} \right]
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| </math>
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| Such an array may be called a "Fibonacci Box". Note that ''''' q', q, p, p' ''''' is a generalized Fibonacci sequence. Taking column, row, and diagonal products we obtain the sides of triangle '''[a, b, c],''' its area '''''A''''', and its perimeter '''''P''''', as well as the radii '''''r'''''<sub>'''''i'''''</sub> of its [[incircle]] and three [[excircles]] as follows:
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| : <math>
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| \begin{array}{l}
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| a = 2qp \\
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| b = q'p' \\
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| c = pp' - qq' = qp' + q'p \\
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| \\
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| \text{radii} \to (r_1 = qq',r_2 = qp',r_3 = q'p,r_4 = pp') \\
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| A = qq'pp' \\
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| P = r_1 + r_2 + r_3 + r_4
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| \end{array}
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| </math>
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| The half-angle tangents at the acute angles are '''''q/p''''' and '''''q'/p''''''.
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| EXAMPLE:
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| Using [[coprime]] integers 9 and 2.
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| : <math>
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| \left[ {\begin{array}{*{20}c}
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| 2 & 9 \\
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| \bullet & \bullet
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| \end{array}} \right] \to \left[ {\begin{array}{*{20}c}
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| 2 & 9 \\
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| 11 & 13
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| \end{array}} \right]
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| </math> | |
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| The column, row, and diagonal products are: (columns: 22 and 117), (rows: 18 and 143), (diagonals: 26 and 99), so
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| : <math>
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| \begin{array}{l}
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| a = 2(22)=44 \\
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| b =117 \\
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| c = (143 - 18) = (26 + 99)=125 \\
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| \\
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| \text{radii} \to (r_1 = 18,\quad r_2 = 26, \quad r_3 = 99 ,\quad r_4 = 143) \\
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| A = (18)(143)=2574 \\
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| P = (18 + 26 + 99 + 143) =286
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| \end{array}
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| </math>
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| The half-angle tangents at the acute angles are 2/11 and 9/13. Note that if the chosen integers '''''q''''', '''''q'''''' are not [[coprime]], the same procedure leads to a non-primitive triple.
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| ====Pythagorean triples and Descartes' circle equation====
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| This method of generating ''primitive Pythagorean triples'' also provides integer solutions to [[Descartes' theorem|Descartes' Circle Equation]],<ref name=Bernhart />
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| : <math>\left( k_1 + k_2 + k_3 + k_4 \right)^2 = 2\left( k_1^2 + k_2^2 + k_3^2 + k_4^2 \right),</math>
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| where integer [[curvature]]s ''k<sub>i</sub>'' are obtained by multiplying the reciprocal of each radius by the area '''''A'''''. The result is ''k''<sub>1</sub> = pp', ''k''<sub>2</sub> = qp', ''k''<sub>3</sub> = q'p, ''k''<sub>4</sub> = qq'. Here, the largest circle is taken as having negative curvature with respect to the other three. The largest circle (curvature ''k''<sub>4</sub>) may also be replaced by a smaller circle with positive curvature ( ''k''<sub>0</sub> = 4''pp' − qq' ''). EXAMPLE: Using the area and four radii obtained above for primitive triple [44, 117, 125] we obtain the following integer solutions to Descartes' Equation: ''k''<sub>1</sub> = 143, ''k''<sub>2</sub> = 99, ''k''<sub>3</sub> = 26, ''k''<sub>4</sub> = (−18), and ''k''<sub>0</sub> = 554.
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| ====A Ternary Tree: Generating All Primitive Pythagorean Triples====
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| Each primitive Pythagorean triple corresponds uniquely to a Fibonacci Box. Conversely, each Fibonacci Box corresponds to a unique and primitive Pythagorean triple. In this section we shall use the Fibonacci Box in place of the primitive triple it represents. An infinite [[ternary tree]] containing all primitive Pythagorean triples/Fibonacci Boxes can be constructed by the following procedure.<ref name=Price>{{cite arxiv|first=H. Lee|last=Price|title=The Pythagorean Tree: A New Species |year=2008|pages=14|eprint=0809.4324|class=math.HO }}</ref>
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| Consider a Fibonacci Box containing two, odd, coprime integers ''x'' and ''y'' in the right-hand column.
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| : <math>
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| \left[ {\begin{array}{*{20}{c}}
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| \bullet & x \\
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| \bullet & y
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| \end{array}} \right]
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| </math>
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| It may be seen that these integers can also be placed as follows:
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|
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| : <math>
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| \left[ {\begin{array}{*{20}{c}}
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| \bullet & x \\
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| y & \bullet
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| \end{array}} \right],\left[ {\begin{array}{*{20}{c}}
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| x & y \\
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| \bullet & \bullet
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| \end{array}} \right],\left[ {\begin{array}{*{20}{c}}
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| y & x \\
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| \bullet & \bullet
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| \end{array}} \right]
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| </math>
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| resulting in three more valid Fibonacci boxes containing ''x'' and ''y''. We may think of the first Box as the “parent” of the next three. For example, if ''x'' = 1 and ''y'' = 3 we have:
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| : <math>
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| \left[ {\begin{array}{*{20}{c}}
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| 1 & 1 \\
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| 2 & 3
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| \end{array}} \right] \leftarrow \text{parent}
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| </math> | |
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| : <math>
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| \left[ {\begin{array}{*{20}{c}}
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| 2 & 1 \\
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| 3 & 5
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| \end{array}} \right],\left[ {\begin{array}{*{20}{c}}
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| 1 & 3 \\
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| 4 & 5
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| \end{array}} \right],\left[ {\begin{array}{*{20}{c}}
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| 3 & 1 \\
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| 4 & 7
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| \end{array}} \right] \leftarrow \text{children}
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| </math>
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| Moreover, each "child" is itself the parent of three more children which can be obtained by the same procedure. Continuing this process at each node leads to an infinite ternary tree containing all possible Fibonacci Boxes, or equivalently, to a ternary tree containing all possible primitive triples. (The tree shown here is distinct from the classic tree described by Berggren in 1934, and has many different number-theoretic properties.) Compare: "Classic Tree".<ref name=Berggren>{{Cite journal|last=Berggren|first=B.|title=Pytagoreiska trianglar| journal=Tidskrift för elementär matematik, fysik och kemi|volume=17|year=1934|pages=129–139|language=Swedish|postscript=.}}</ref> See also [[Tree of primitive Pythagorean triples]].
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| ==Generating triples using quadratic equations==
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| There are several methods for defining [[quadratic equations]] for calculating each leg of a Pythagorean triple.<ref>J. L. Poet and D. L. Vestal, Jr. (2005). “Curious Consequences of a Miscopied Quadratic, ” ''College Mathematics Journal'' 36 , 273–277.</ref> A simple method is to modify the standard Euclid equation by adding a variable ''x'' to each ''m'' and ''n'' pair. The ''m, n'' pair is treated as a constant while the value of ''x'' is varied to produce a “family” of triples based on the selected triple. An arbitrary coefficient can be placed in front of the “''x''” value on either ''m'' or ''n'', which causes the resulting equation to systematically “skip” through the triples. For example, let’s use the triple [20, 21, 29] which can be calculated from the Euclid equations with a value of ''m'' = 5 and ''n'' = 2. Also, let’s arbitrarily put the coefficient of 4 in front of the “''x''” in the “''m''” term.
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| Let <math>m_1=(4x+m)</math> and let <math>n_1=(x+n)</math>
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| Hence, substituting the values of ''m'' and ''n'':
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| : <math>\begin{align}
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| \text{Side }A & =2m_1 n_1 & & = 2(4x+5)\text{ }(x+2) & & = 8x^2+26x+20 \\
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| \text{Side }B & =m_1^2-n_1^2 & & = (4x+5)^2-(x+2)^2 & & = 15x^2+36x+21 \\
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| \text{Side }C & =m_1^2+n_1^2 & & = (4x+5)^2+(x+2)^2 & & = 17x^2+44x+29
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| \end{align}</math>
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| | |
| Note that the original triple comprises the constant term in each of the respective quadratic equations. Below is a sample output from these equations. Note that the effect of these equations is to cause the “''m''” value in the Euclid equations to increment in steps of 4, while the “''n''” value increments by 1.
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| <span class="wikiEditor-tab"></span>
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| x side a side b side c m n
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| 0 20 21 29 5 2
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| 1 54 72 90 9 3
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| 2 104 153 185 13 4
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| 3 170 264 314 17 5
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| 4 252 405 477 21 6
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| | |
| ==Pythagorean triples by use of matrices and linear transformations==
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| | |
| {{main|Tree of primitive Pythagorean triples}}
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| | |
| Let <math>[\text{ }a,\text{ }b,\text{ }c]</math> be a primitive triple with <math>a</math> odd. Then 3 new triples <math>[{{a}_{1}},{{b}_{1}},{{c}_{1}}], \text{ }[{{a}_{2}},{{b}_{2}},{{c}_{2}}], \text{ }[{{a}_{3}},{{b}_{3}},{{c}_{3}}]</math> may be produced from <math>[\text{ }a,\text{ }b,\text{ }c]</math> using [[matrix multiplication]] and Berggren’s<ref name="Berggren"/> three matrices ''A, B, C''. Triple <math>[\text{ }a,\text{ }b,\text{ }c]</math> is termed the "''parent''" of the three new triples (the "''children''"). Each child is itself the parent of 3 more children, and so on. If one begins with primitive triple <math>[\text{ }3,\text{ }4,\text{ }5]</math>, all primitive triples will eventually be produced by application of these matrices. The result can be graphically represented as an infinite [[ternary tree]] with <math>[\text{ }a,\text{ }b,\text{ }c]</math> at the root node. An equivalent result may be obtained using Berggrens's three [[linear transformations]] shown below.
| |
| | |
| : <math>\overset{A}{\mathop{\left[ \begin{matrix}
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| -1 & 2 & 2 \\
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| -2 & 1 & 2 \\
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| -2 & 2 & 3 \\
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| \end{matrix} \right]}} \left[ \begin{matrix}
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| a \\
| |
| b \\
| |
| c \\
| |
| \end{matrix} \right]=\left[ \begin{matrix}
| |
| a_1 \\
| |
| b_1 \\
| |
| c_1 \\
| |
| \end{matrix} \right],\quad \text{ }\overset{B}{\mathop{\left[ \begin{matrix}
| |
| 1 & 2 & 2 \\
| |
| 2 & 1 & 2 \\
| |
| 2 & 2 & 3 \\
| |
| \end{matrix} \right]}} \left[ \begin{matrix}
| |
| a \\
| |
| b \\
| |
| c \\
| |
| \end{matrix} \right]=\left[ \begin{matrix}
| |
| a_2 \\
| |
| b_2 \\
| |
| c_2
| |
| \end{matrix} \right],\quad \text{ }\overset{C}{\mathop{\left[ \begin{matrix}
| |
| 1 & -2 & 2 \\
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| 2 & -1 & 2 \\
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| 2 & -2 & 3
| |
| \end{matrix} \right]}} \left[ \begin{matrix}
| |
| a \\
| |
| b \\
| |
| c
| |
| \end{matrix} \right]=\left[ \begin{matrix}
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| a_3 \\
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| b_3 \\
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| c_3
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| \end{matrix} \right]</math>
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| | |
| Berggren's three linear transformations are:
| |
| | |
| : <math>\begin{align}
| |
| & \begin{matrix}
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| -a+2b+2c=a_1 \quad & -2a+b+2c=b_1 \quad & -2a+2b+3c=c_1 & \quad\to \left[ \text{ }a_1,\text{ }b_1,\text{ }c_1 \right] \\
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| \end{matrix} \\
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| & \begin{matrix}
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| +a+2b+2c={{a}_{2}} \quad & +2a+b+2c={{b}_{2}} \quad & +2a+2b+3c={{c}_{2}} & \quad\to \left[ \text{ }{{a}_{2}},\text{ }{{b}_{2}},\text{ }{{c}_{2}} \right] \\
| |
| \end{matrix} \\
| |
| & \begin{matrix}
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| +a-2b+2c={{a}_{3}} \quad & +2a-b+2c={{b}_{3}} \quad & +2a-2b+3c={{c}_{3}} & \quad\to \left[ \text{ }{{a}_{3}},\text{ }{{b}_{3}},\text{ }{{c}_{3}}\right] \\
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| \end{matrix} \\
| |
| &
| |
| \end{align}</math>
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| | |
| Alternatively, one may also use 3 different matrices found by Price.<ref name="Price"/> These matrices ''A', B', C''' and their corresponding linear transformations are shown below.
| |
| | |
| : <math>\overset{{{A}'}}{\mathop{\left[ \begin{matrix}
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| 2 & 1 & -1 \\
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| -2 & 2 & 2 \\
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| -2 & 1 & 3
| |
| \end{matrix} \right]}} \left[ \begin{matrix}
| |
| a \\
| |
| b \\
| |
| c
| |
| \end{matrix} \right]=\left[ \begin{matrix}
| |
| a_1 \\
| |
| b_1 \\
| |
| c_1
| |
| \end{matrix} \right],\quad \text{ }\overset{{{B}'}}{\mathop{\left[ \begin{matrix}
| |
| 2 & 1 & 1 \\
| |
| 2 & -2 & 2 \\
| |
| 2 & -1 & 3
| |
| \end{matrix} \right]}} \left[ \begin{matrix}
| |
| a \\
| |
| b \\
| |
| c \\
| |
| \end{matrix} \right]=\left[ \begin{matrix}
| |
| a_2 \\
| |
| b_2 \\
| |
| c_2
| |
| \end{matrix} \right],\quad \text{ }\overset{{{C}'}}{\mathop{\left[ \begin{matrix}
| |
| 2 & -1 & 1 \\
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| 2 & 2 & 2 \\
| |
| 2 & 1 & 3 \\
| |
| \end{matrix} \right]}} \left[ \begin{matrix}
| |
| a \\
| |
| b \\
| |
| c \\
| |
| \end{matrix} \right]=\left[ \begin{matrix}
| |
| a_3 \\
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| b_3 \\
| |
| c_3
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| \end{matrix} \right]</math>
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| | |
| Price's three linear transformations are:
| |
| | |
| : <math>\begin{align}
| |
| & \begin{matrix}
| |
| +2a+b-c=a_1 \quad & -2a+2b+2c=b_1 \quad & -2a+b+3c=c_1 & \quad \to \left[ \text{ }a_1,\text{ }b_1,\text{ }c_1 \right]
| |
| \end{matrix} \\
| |
| & \begin{matrix}
| |
| +2a+b+c=a_2 \quad & +2a-2b+2c=b_2 \quad & +2a-b+3c=c_2 & \quad \to \left[ \text{ }a_2,\text{ }b_2,\text{ }c_2 \right]
| |
| \end{matrix} \\
| |
| & \begin{matrix}
| |
| +2a-b+c=a_3 \quad & +2a+2b+2c=b_3 \quad & +2a+b+3c=c_3 & \quad \to \left[ \text{ }a_3,\text{ }b_3,\text{ }c_3 \right]
| |
| \end{matrix} \\
| |
| &
| |
| \end{align}</math>
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| | |
| The "3 children" produced by each of the two sets of matrices are not the same, but each set separately produces all primitive triples.
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|
| |
| EXAMPLE: Using [ 5, 12, 13] as the parent, we get two sets of three children:
| |
| | |
| : <math>\begin{matrix}
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| {} & \left[ \text{5},12,13 \right] & {} \\
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| A & B & C \\
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| \left[ 45,28,53 \right] & \left[ \text{55,48,73} \right] & \left[ \text{7,24,25} \right]
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| \end{matrix}\quad \quad \quad \quad \quad \quad \begin{matrix}
| |
| {} & \left[ \text{5},12,13 \right] & {} \\
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| {{A}'} & {{B}'} & {{C}'} \\
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| \left[ 9,40,41 \right] & \left[ \text{35,12,37} \right] & \left[ \text{11,60,61} \right]
| |
| \end{matrix}</math>
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| | |
| ==Area proportional to sums of squares==
| |
| | |
| All primitive triples with <math>b+1=c</math> and with ''a'' odd can be generated as follows:<ref>Barbeau, Edward, ''Power Play'', Mathematical Association of America,1997, p. 51, item 3.</ref>
| |
| | |
| {| class="wikitable"
| |
| |-
| |
| ! Pythagorean triple !! Semi-perimeter !! Area !! Incircle radius !! Circumcircle radius
| |
| |-
| |
| | <math>\left( 3,4,5 \right)</math> || 1 + 2 + 3 || <math>6\times (1^2)</math> || 1 || <math>\tfrac{5}{2}</math>
| |
| |-
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| | <math>\left( 5,12,13 \right)</math> || 1 + 2 + 3 + 4 + 5 || <math>6\times (1^2+2^2)</math> || 2 || <math>\tfrac{13}{2}</math>
| |
| |-
| |
| | <math>\left( 7,24,25 \right)</math> || 1 + 2 + 3 + 4 + 5 + 6 + 7 || <math>6\times (1^2+2^2+3^2)</math> || 3 || <math>\tfrac{25}{2}</math>
| |
| |-
| |
| | ....... || ....... || ....... || ....... || .......
| |
| |-
| |
| | <math>\left( a,\tfrac{a^2-1}{2},\tfrac{a^2+1}{2} \right)</math> || 1 + 2 + ... + ''a'' || <math>6\times \left[ 1^2+2^2+\cdots+\left( \tfrac{a-1}{2} \right)^2 \right]</math> || <math>\left( \tfrac{a-1}{2} \right)</math> || <math>\tfrac{c}{2}</math>
| |
| |}
| |
| | |
| ==Height-excess enumeration theorem==
| |
| | |
| Wade and Wade<ref>Wade, Peter, and Wade, William, "Recursions that produce Pythoagorean triples", ''[[College Mathematics Journal]]'' 31, March 2000, 98-101.</ref> first introduced the categorization of Pythagorean triples by their height, defined as c - b, linking 3,4,5 to 5,12,13 and 7,24,25 and so on.
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| | |
| McCullough and Wade<ref>McCullough, Darryl, and Wade, Elizabeth, "Recursive enumeration of Pythagorean triples", ''[[College Mathematics Journal]]'' 34, March 2003, 107-111.</ref> extended this approach, which produces all Pythagorean triples when <math>k> \frac{h \sqrt{2}}{d}:</math> Write a positive integer ''h'' as pq<sup>2</sup> with ''p'' square-free and ''q'' positive. Set ''d'' = 2''pq'' if ''p'' is odd, or ''d''= ''pq'' if ''p'' is even. For all pairs (''h,k'') of positive integers, the triples are given by
| |
| | |
| :<math>(h+dk, dk+\frac{(dk)^2}{2h}, h+dk+\frac{(dk)^2}{2h}).</math>
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| The primitive triples occur when gcd(''k, h'') = 1 and either ''h=q''<sup>2</sup> with ''q'' odd or ''h''=2''q''<sup>2</sup>.
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| ==References==
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| {{reflist}}
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| {{DEFAULTSORT:Formulas For Generating Pythagorean Triples}}
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| [[Category:Number theory]]
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