Pythagorean triple

Revision as of 19:09, 6 March 2014 by Greenpepper9999 (talk | contribs) (Common Pythagorean Triples)

A Pythagorean triple is a triple of positive integers, $(a, b, c)$ such that $a^2 + b^2 = c^2$. Pythagorean triples arise in geometry as the side-lengths of right triangles.

Common Pythagorean Triples

These are some common Pythagorean triples: *=Primitive (see below)

(3, 4, 5)*

(5, 12, 13)*

(6, 8, 10)

(7, 24, 25)*

(8, 15, 17) *

(9, 12, 15)

(9, 40, 41) *

(20, 21, 29)*

(11, 60, 61)*

(13, 84, 85)*

(12, 35, 37) *

(16, 63, 65) *

(36, 77, 85)*

(33, 56, 65) *

(39, 80, 89)*

(28, 45, 53)*

(48, 55, 73) *

(65, 72, 97)*

Pythagorean Triangles

Each positive integer solution of the diophantine equation $a^2 + b^2 = c^2$ defining the Pythagorean triples satisfies $a, b < c < a + b$. Thus, any triple of positive integers satisfying this equation also satisfies the triangle inequality, so the solutions correspond to right triangles with integral side lengths.

Primitive Pythagorean Triples

A Pythagorean triple is called primitive if its three members have no common divisors, so that they are relatively prime. All triples listed above are primitive. Integral multiples of Pythagorean triples will also satisfy $a^2 + b^2 = c^2$, but they will not form primitive triples. For example, all triples of integers of the form $(3n, 4n, 5n)$, such as $(6, 8, 10)$, are Pythagorean triples.

General Form of Primitive Pythagorean Triples

Theorem. A triple of integers is a primitive Pythagorean triple if and only if it may be written in the form $(m^2-n^2, 2mn, m^2 + n^2)$ or $(2mn, m^2-n^2, m^2+n^2)$, where $m > n$ are relatively prime positive integers of different parity.

Proof

Let $(a,b,c)$ be a primitive Pythagorean triple. If $a$ and $b$ both odd, then we must have \[c^2 \equiv a^2 + b^2 \equiv 1+1 \equiv 2 \pmod{4},\] which is a contradiction, since 2 is not a square mod 4. Hence at least one of $a$ and $b$, say $b$, is even. Then $a$ must be odd, since $a$ and $b$ must be relatively prime. It follows that $c$ is odd as well. It follows that the numbers $s = (c+a)/2$ and $d = (c-a)/2$ are positive integers. These positive integers must be relatively prime, since any common divisor of $s$ and $d$ must divide both $s+d= c$ and $s-d = a$. Since $a = s-d$ and $c= s+d$, it follows that \[b = \sqrt{c^2 -a^2} = \sqrt{(s+d)^2 - (s-d)^2} = \sqrt{4sd} = 2\sqrt{sd} .\] Since $\sqrt{sd} = b/2$ must be an integer and $s$ and $d$ are relatively prime, it follows that $s$ and $d$ are perfect squares. Hence we may denote $s= m^2$ and $d = n^2$ for integers $m$ and $n$. Since $m^2 + n^2 = s+d = c$ is odd, it follows that $m^2$ and $n^2$ must have different parity, so $m$ and $n$ have different parity. Finally, we observe that \[(m^2-n^2)^2 + (2mn)^2 = m^4 + 2m^2n^2 + n^4 = (m^2+n^2)^2,\] so any triple of the form specified in the theorem is a Pythagorean triple; it must furthermore be a primitive Pythagorean triple, since any common factor of $m^2 - n^2$ and $m^2 + n^2$ (both of which are odd integers, since $m$ and $n$ have different parity) must also be a factor of both $(m^2 + n^2) + (m^2-n^2) = 2m^2$ and $(m^2+n^2) - (m^2 - n^2) = 2n^2$, which are integers with no common factor greater than 2. $\blacksquare$

See also

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