Editing Pythagorean triple (section)

===Proof of Euclid's formula===

That satisfaction of Euclid's formula by ''a, b, c'' is [[Necessary and sufficient condition|sufficient]] for the triangle to be Pythagorean is apparent from the fact that for positive integers {{math|''m''}} and {{math|''n''}}, {{math|''m'' > ''n''}}, the {{math|''a''}}, {{math|''b''}}, and {{math|''c''}} given by the formula are all positive integers, and from the fact that

:<math> a^2+b^2 = (m^2 - n^2)^2 + (2mn)^2 = (m^2 + n^2)^2 = c^2. </math>

A proof of the ''necessity'' that ''a, b, c'' be expressed by Euclid's formula for any primitive Pythagorean triple is as follows.<ref>{{citation |last1=Beauregard |first1=Raymond A. |last2=Suryanarayan |first2=E. R. |chapter=Parametric representation of primitive Pythagorean triples |editor-first=Roger B. |editor-last=Nelsen |title=Proofs Without Words: More Exercises in Visual Thinking |year=2000 |publisher=[[Mathematical Association of America]] |isbn=978-0-88385-721-2 |volume=II |oclc=807785075 |page=[https://archive.org/details/proofswithoutwor0000nels/page/120 120] |chapter-url=https://archive.org/details/proofswithoutwor0000nels/page/120 }}</ref> All such primitive triples can be written as {{math|(''a'', ''b'', ''c'')}} where {{math|1=''a''{{sup|2}} + ''b''{{sup|2}} = ''c''{{sup|2}}}} and {{math|''a''}}, {{math|''b''}}, {{math|''c''}} are [[coprime]]. Thus {{math|''a''}}, {{math|''b''}}, {{math|''c''}} are [[pairwise coprime]] (if a prime number divided two of them, it would be forced also to divide the third one). As {{math|''a''}} and {{math|''b''}} are coprime, at least one of them is odd. If we suppose that {{math|''a''}} is odd, then {{math|''b''}} is even and {{math|''c''}} is odd (if both {{math|''a''}} and {{math|''b''}} were odd, {{math|''c''}} would be even, and {{math|''c''{{sup|2}}}} would be a multiple of 4, while {{math|''a''{{sup|2}} + ''b''{{sup|2}}}} would be [[modular arithmetic|congruent to 2 modulo 4]], as an odd square is congruent to 1 modulo 4).

From <math>a^2+b^2=c^2,</math> assume {{math|''a''}} is odd. We obtain <math>c^2-a^2=b^2</math> and hence <math>(c-a)(c+a)=b^2.</math> Then <math>\tfrac{(c+a)}{b}=\tfrac{b}{(c-a)}.</math> Since <math>\tfrac{(c+a)}{b}</math> is rational, we set it equal to <math>\tfrac{m}{n}</math> in lowest terms. Thus <math>\tfrac{(c-a)}{b}=\tfrac{n}{m},</math> being the reciprocal of  <math>\tfrac{(c+a)}{b}.</math> Then solving

:<math>\frac{c}{b}+\frac{a}{b}=\frac{m}{n}, \quad \quad  \frac{c}{b}-\frac{a}{b}=\frac{n}{m}</math>

for <math>\tfrac{c}{b}</math> and <math>\tfrac{a}{b}</math> gives

:<math>\frac{c}{b}=\frac{1}{2}\left(\frac{m}{n}+\frac{n}{m}\right)=\frac{m^2+n^2}{2mn}, \quad \quad \frac{a}{b}=\frac{1}{2}\left(\frac{m}{n}-\frac{n}{m}\right)=\frac{m^2-n^2}{2mn}.</math>

As <math>\tfrac{m}{n}</math> is fully reduced, {{math|''m''}} and {{math|''n''}} are coprime, and they cannot both be even. If they were both odd, the numerator of <math>\tfrac{m^2-n^2}{2mn}</math> would be a multiple of 4 (because an odd square is congruent to 1 modulo 4), and the denominator 2''mn'' would not be a multiple of 4. Since 4 would be the minimum possible even factor in the numerator and 2 would be the maximum possible even factor in the denominator, this would imply {{math|''a''}} to be even despite defining it as odd. Thus one of {{math|''m''}} and {{math|''n''}} is odd and the other is even, and the numerators of the two fractions with denominator 2''mn'' are odd. Thus these fractions are fully reduced (an odd prime dividing this denominator divides one of {{math|''m''}} and {{math|''n''}} but not the other; thus it does not divide {{math|''m''{{sup|2}} ± ''n''{{sup|2}}}}). One may thus equate numerators with numerators and denominators with denominators, giving Euclid's formula 
:<math> a = m^2 - n^2   ,\ \, b = 2mn ,\ \, c = m^2 + n^2</math> with {{math|''m''}} and {{math|''n''}} coprime and of opposite parities.

A longer but more commonplace proof is given in Maor (2007)<ref>[[Eli Maor|Maor, Eli]], ''The Pythagorean Theorem'', Princeton University Press, 2007: Appendix B.</ref> and Sierpiński (2003).<ref name=Sierpinski>{{citation |author-link=Waclaw Sierpinski |first=Wacław |last=Sierpiński |title=Pythagorean Triangles|title-link=Pythagorean Triangles |at=[https://books.google.com/books?id=6vOfpjmCd7sC&pg=PR4 pp. iv–vii] |year=2003 |publisher=Dover |isbn=978-0-486-43278-6  }}</ref> Another proof is given in {{slink|Diophantine equation|Example of Pythagorean triples}}, as an instance of a general method that applies to every [[homogeneous polynomial|homogeneous]] Diophantine equation of degree two.