2000 | OriginalPaper | Chapter
Pythagorean Triples
Author : Edmund Hlawka
Published in: Number Theory
Publisher: Hindustan Book Agency
Included in: Professional Book Archive
Activate our intelligent search to find suitable subject content or patents.
Select sections of text to find matching patents with Artificial Intelligence. powered by
Select sections of text to find additional relevant content using AI-assisted search. powered by
The investigation of Pythagorian triples has a very long history. For the first hundred years I refer to the famous book [DIC01]. Triangles of this type were given by Greek and Indian mathematicians. Arithmetically these are the solutions of the diophantine equation <math display='block'> <mrow> <msup> <mi>x</mi> <mn>2</mn> </msup> <mo>+</mo><msup> <mi>y</mi> <mn>2</mn> </msup> <mo>=</mo><msup> <mi>z</mi> <mn>2</mn> </msup> </mrow> </math> $${x^2} + {y^2} = {z^2}$$ in rational numbers. The general solution is given by the formulas <math display='block'> <math display='block'> <mrow> <mtable columnalign='left'> <mtr columnalign='left'> <mtd columnalign='left'> <mrow> <mi>x</mi><mo>=</mo><mi>l</mi><mo stretchy='false'>(</mo><msup> <mi>m</mi> <mn>2</mn> </msup> <mo>−</mo><msup> <mi>n</mi> <mn>2</mn> </msup> <mo stretchy='false'>)</mo> </mrow> </mtd> </mtr> <mtr columnalign='left'> <mtd columnalign='left'> <mrow> <mi>y</mi><mo>=</mo><mi>l</mi><mo>⋅</mo><mn>2</mn><mi>m</mi><mi>n</mi> </mrow> </mtd> </mtr> <mtr columnalign='left'> <mtd columnalign='left'> <mrow> <mi>z</mi><mo>=</mo><mi>l</mi><mo stretchy='false'>(</mo><msup> <mi>m</mi> <mn>2</mn> </msup> <mo>+</mo><msup> <mi>n</mi> <mn>2</mn> </msup> <mo stretchy='false'>)</mo><mtext> </mtext><mi>l</mi><mo stretchy='false'>(</mo><mi>l</mi><mo>≠</mo><mn>0</mn><mo stretchy='false'>)</mo><mo>,</mo><mtext> </mtext><mi>m</mi><mo>,</mo><mi>n</mi><mo>,</mo><mtext> </mtext><mi>a</mi><mi>r</mi><mi>b</mi><mi>i</mi><mi>t</mi><mi>r</mi><mi>a</mi><mi>r</mi><mi>y</mi><mo>.</mo> </mrow> </mtd> </mtr> </mtable> </mrow> </math>$$\begin{array}{*{20}{l}} {x = l({m^2} - {n^2})} \\ {y = l \cdot 2mn} \\ {z = l({m^2} + {n^2})\;l(l \ne 0),\;m,n,\,arbitrary.} \end{array}$$.