Editing Pythagorean triple (section)

==Special cases and related equations==

===The Platonic sequence===
The case {{math|1=''n'' = 1}} of the more general construction of Pythagorean triples has been known for a long time. [[Proclus]], in his commentary to the [[Pythagorean theorem|47th Proposition]] of the first book of [[Euclid's Elements|Euclid's ''Elements'']], describes it as follows:

<blockquote>Certain methods for the discovery of triangles of this kind are handed down, one which they refer to Plato, and another to [[Pythagoras]]. (The latter) starts from odd numbers. For it makes the odd number the smaller of the sides about the right angle; then it takes the square of it, subtracts unity and makes half the difference the greater of the sides about the right angle; lastly it adds unity to this and so forms the remaining side, the hypotenuse.<br />
...For the method of Plato argues from even numbers. It takes the given even number and makes it one of the sides about the right angle; then, bisecting this number and squaring the half, it adds unity to the square to form the hypotenuse, and subtracts unity from the square to form the other side about the right angle. ... Thus it has formed the same triangle that which was obtained by the other method.</blockquote>

In equation form, this becomes:

{{math|''a''}} is odd (Pythagoras, c. 540 BC):

:<math>\text{side }a : \text{side }b = {a^2 - 1 \over 2} : \text{side }c = {a^2 + 1 \over 2}.</math>

{{math|''a''}} is even (Plato, c. 380 BC):

:<math>\text{side }a : \text{side }b = \left({a \over 2}\right)^2 - 1 : \text{side }c = \left({a \over 2}\right)^2 + 1</math>

It can be shown that all Pythagorean triples can be obtained, with appropriate rescaling, from the basic Platonic sequence ({{math|''a''}}, {{math|(''a''{{sup|2}} − 1)/2}} and {{math|(''a''{{sup|2}} + 1)/2}}) by allowing {{math|''a''}} to take non-integer rational values. If {{math|''a''}} is replaced with the fraction {{math|''m''/''n''}} in the sequence, the result is equal to the 'standard' triple generator (2''mn'', {{math|''m''{{sup|2}} − ''n''{{sup|2}}}},{{math|''m''{{sup|2}} + ''n''{{sup|2}}}}) after rescaling. It follows that every triple has a corresponding rational {{math|''a''}} value which can be used to generate a [[similarity (geometry)|similar]] triangle (one with the same three angles and with sides in the same proportions as the original).  For example, the Platonic equivalent of {{math|1=(56, 33, 65)}} is generated by {{math|1=''a'' = ''m''/''n'' = 7/4}} as {{math|1=(''a'', (''a''{{sup|2}} –1)/2, (''a''{{sup|2}}+1)/2) = (56/32, 33/32, 65/32)}}. The Platonic sequence itself can be derived{{clarify|What does this mean?|date=April 2015}} by following the steps for 'splitting the square' described in [[Diophantus II.VIII]].

===The Jacobi–Madden equation===
{{Main|Jacobi–Madden equation}}
The equation,

:<math>a^4+b^4+c^4+d^4 = (a+b+c+d)^4</math>

is equivalent to the special Pythagorean triple,

:<math>(a^2+ab+b^2)^2+(c^2+cd+d^2)^2 = ((a+b)^2+(a+b)(c+d)+(c+d)^2)^2</math>

There is an infinite number of solutions to this equation as solving for the variables involves an [[elliptic curve]]. Small ones are,

:<math>a, b, c, d = -2634, 955, 1770, 5400</math>

:<math>a, b, c, d = -31764, 7590, 27385, 48150</math>

===Equal sums of two squares===

One way to generate solutions to <math>a^2+b^2=c^2+d^2</math> is to parametrize ''a, b, c, d''  in terms of integers ''m, n, p, q'' as follows:<ref>{{citation
 | last = Nahin | first = Paul J. | author-link = Paul J. Nahin
 | isbn = 0-691-02795-1
 | location = Princeton, New Jersey
 | mr = 1645703
 | pages = 25–26
 | publisher = Princeton University Press
 | title = An Imaginary Tale: The Story of <math>\sqrt{-1}</math>
 | year = 1998}}</ref>

:<math>(m^2+n^2)(p^2+q^2)=(mp-nq)^2+(np+mq)^2=(mp+nq)^2+(np-mq)^2.</math>

===Equal sums of two fourth powers===

Given two sets of Pythagorean triples,

:<math>(a^2-b^2)^2+(2a b)^2 = (a^2+b^2)^2</math>

:<math>(c^2-d^2)^2+(2c d)^2 = (c^2+d^2)^2</math>

the problem of finding equal products of a [[Cathetus|non-hypotenuse side]] and the hypotenuse,

:<math>(a^2 -b^2)(a^2+b^2) = (c^2 -d^2)(c^2+d^2)</math>

is easily seen to be equivalent to the equation,

:<math>a^4 -b^4 = c^4 -d^4</math>

and was first solved by Euler as <math>a, b, c, d = 133,59,158,134.</math>  Since he showed this is a rational point in an [[elliptic curve]], then there is an infinite number of solutions. In fact, he also found a 7th degree polynomial parameterization.

===Descartes' Circle Theorem===

For the case of [[Descartes' theorem|Descartes' circle theorem]] where all variables are squares,

:<math>2(a^4+b^4+c^4+d^4) = (a^2+b^2+c^2+d^2)^2</math>

Euler showed this is equivalent to three simultaneous Pythagorean triples,

:<math>(2ab)^2+(2cd)^2 = (a^2+b^2-c^2-d^2)^2</math>

:<math>(2ac)^2+(2bd)^2 = (a^2-b^2+c^2-d^2)^2</math>

:<math>(2ad)^2+(2bc)^2 = (a^2-b^2-c^2+d^2)^2</math>

There is also an infinite number of solutions, and for the special case when <math>a+b=c</math>, then the equation simplifies to,

:<math>4(a^2+a b+b^2) = d^2</math>

with small solutions as <math>a, b, c, d = 3, 5, 8, 14</math> and can be solved as [[quadratic forms|binary quadratic forms]].

===Almost-isosceles Pythagorean triples===

No Pythagorean triples are [[isosceles]], because the ratio of the hypotenuse to either other side is {{radic|2}}, but [[Square root of 2#Proofs of irrationality|{{radic|2}} cannot be expressed as the ratio of 2 integers]].

There are, however, [[Special right triangles#Almost-isosceles Pythagorean triples|right-angled triangles]] with integral sides for which the lengths of the [[Cathetus|non-hypotenuse sides]] differ by one, such as,

:<math>3^2+4^2 = 5^2</math>

:<math>20^2+21^2 = 29^2</math>

and an infinite number of others. They can be completely parameterized as,

:<math>\left(\tfrac{x-1}{2}\right)^2+\left(\tfrac{x+1}{2}\right)^2 = y^2</math>

where {''x, y''} are the solutions to the [[Pell equation]] <math>x^2-2y^2 = -1.</math>

If {{math|''a''}}, {{math|''b''}}, {{math|''c''}} are the sides of this type of primitive Pythagorean triple then the solution to the Pell equation is given by the [[Recurrence relation|recursive formula]]

:<math>a_n=6a_{n-1}-a_{n-2}+2</math> with <math>a_1=3</math> and <math>a_2=20</math>
:<math>b_n=6b_{n-1}-b_{n-2}-2</math> with <math>b_1=4</math> and <math>b_2=21</math>
:<math>c_n=6c_{n-1}-c_{n-2}</math> with <math>c_1=5</math> and {{tmath|1=c_2=29}}.<ref>{{cite OEIS |A001652 |mode=cs2}}; {{cite OEIS |A001653 |mode=cs2}}</ref>

This sequence of primitive Pythagorean triples forms the central stem (trunk) of the [[Tree of primitive Pythagorean triples|rooted ternary tree]] of primitive Pythagorean triples.

When it is the longer non-hypotenuse side and hypotenuse that differ by one, such as in

:<math>5^2+12^2 = 13^2</math>

:<math>7^2+24^2 = 25^2</math>

then the complete solution for the primitive Pythagorean triple {{math|''a''}}, {{math|''b''}}, {{math|''c''}} is

:<math>a=2m+1, \quad b=2m^2+2m, \quad c=2m^2+2m+1</math>

and

:<math>(2m+1)^2+(2m^2+2m)^2=(2m^2+2m+1)^2</math>

where integer <math>m>0</math> is the generating parameter.

It shows that all [[odd numbers]] (greater than 1) appear in this type of almost-isosceles primitive Pythagorean triple. This sequence of primitive Pythagorean triples forms the right hand side outer stem of the rooted ternary tree of primitive Pythagorean triples.

Another property of this type of almost-isosceles primitive Pythagorean triple is that the sides are related such that 
:<math>a^b+b^a=Kc</math>
for some integer <math>K</math>. Or in other words <math>a^b+b^a</math> is divisible by <math>c</math> such as in
:<math>(5^{12}+12^5)/13 = 18799189</math>.<ref>{{cite OEIS|A303734|mode=cs2}}</ref>

===Fibonacci numbers in Pythagorean triples===

Starting with 5, every second [[Fibonacci number]] is the length of the hypotenuse of a right triangle with integer sides, or in other words, the largest number in a Pythagorean triple, obtained from the formula
<math display=block>(F_nF_{n+3})^2 + (2F_{n+1}F_{n+2})^2 = F_{2n+3}^2.</math>
The sequence of Pythagorean triangles obtained from this formula has sides of lengths
:(3,4,5), (5,12,13), (16,30,34), (39,80,89), ...
The middle side of each of these triangles is the sum of the three sides of the preceding triangle.<ref>{{citation
 | last = Pagni | first = David
 | date = September 2001
 | issue = 4
 | journal = Mathematics in School
 | jstor = 30215477
 | pages = 39–40
 | title = Fibonacci Meets Pythagoras
 | volume = 30}}</ref>