Difference between revisions of "Pythagorean triple"

Revision as of 22:53, 18 December 2007

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:

(3, 4, 5)

(20, 21, 29)

(11, 60, 61)

(13, 84, 85)

(5, 12, 13)

(12, 35, 37)

(16, 63, 65)

(36, 77, 85)

(8, 15, 17)

(9, 40, 41)

(33, 56, 65)

(39, 80, 89)

(7, 24, 25)

(28, 45, 53)

(48, 55, 73)

(65, 72, 97)

Primitive Pythagorean Triples

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

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. $\blacksquare$

@@ Line 38: / Line 38: @@
 == Primitive Pythagorean Triples ==
 A Pythagorean triple is called ''primitive'' if its three members have no common [[divisor]]s, so that they are [[relatively prime]].  All of the above triples are primitive.  Integral [[multiple]]s of the above triples will also satisfy <math>a^2 + b^2 = c^2</math>, but they will not form primitive triples.  For example, any three numbers in the form of <math>(3x, 4x, 5x)</math>, such as <math>(6, 8, 10)</math>, will also satisfy it.
+=== 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 <math>(m^2-n^2, 2mn, m^2 + n^2)</math> or <math>(2mn, m^2-n^2, m^2+n^2)</math>, where <math>m > n</math> are relatively prime positive integers of different [[parity]].
+==== Proof ====
+Let <math>(a,b,c)</math> be a primitive Pythagorean triple.  If <math>a</math> and <math>b</math> both odd, then we must have
+<cmath> c^2 \equiv a^2 + b^2 \equiv 1+1 \equiv 2 \pmod{4}, </cmath>
+which is a contradiction, since 2 is not a [[quadratic residue | square]] [[modulus | mod]] 4.  Hence at least one of <math>a</math> and <math>b</math>, say <math>b</math>, is even.  Then <math>a</math> must be odd, since <math>a</math> and <math>b</math> must be relatively prime.  It follows that <math>c</math> is odd as well.  It follows that the numbers <math>s = (c+a)/2</math> and <math>d = (c-a)/2</math> are positive integers.  These positive integers must be relatively prime, since any common divisor of <math>s</math> and <math>d</math> must divide both <math>s+d= c</math> and <math>s-d = a</math>.  Since <math>a = s-d</math> and <math>c= s+d</math>, it follows that
+<cmath> b = \sqrt{c^2 -a^2} = \sqrt{(s+d)^2 - (s-d)^2} = \sqrt{4sd} = 2\sqrt{sd} . </cmath>
+Since <math>\sqrt{sd} = b/2</math> must be an integer and <math>s</math> and <math>d</math> are relatively prime, it follows that <math>s</math> and <math>d</math> are perfect squares.  Hence we may denote <math>s= m^2</math> and <math>d = n^2</math> for integers <math>m</math> and <math>n</math>.  Since <math>m^2 + n^2 = s+d = c</math> is odd, it follows that <math>m^2</math> and <math>n^2</math> must have different parity, so <math>m</math> and <math>n</math> have different parity.  Finally, we observe that
+<cmath> (m^2-n^2)^2 + (2mn)^2 = m^4 + 2m^2n^2 + n^4 = (m^2+n^2)^2, </cmath>
+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 <math>m^2 - n^2</math> and <math>m^2 + n^2</math> (both of which are odd integers, since <math>m</math> and <math>n</math> have different parity) must also be a factor of both <math>(m^2 + n^2) + (m^2-n^2) = 2m^2</math> and <math>(m^2+n^2) - (m^2 - n^2) = 2n^2</math>, which are integers with no common factor greater than 2.  <math>\blacksquare</math>
 == See also ==
 * [[Pythagorean Theorem]]
 * [[Diophantine equation]]

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