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A Pythagorean triple consists of three positive integers a, b, and c, such that a2 + b2 = c2. Such a triple is commonly written (a, b, c), and a well-known example is (3, 4, 5). If (a, b, c) is a Pythagorean triple, then so is (ka, kb, kc) for any positive integer k. A primitive Pythagorean triple is one in which a, b and c are coprime.

The name is derived from the Pythagorean theorem, of which every Pythagorean triple is a solution. The converse is not true. For instance, the triangle with sides a = b = 1 and c = &radic;2 is right, but (1, 1, &radic;2) is not a Pythagorean triple because &radic;2 is not an integer. Moreover, 1 and &radic;2 don't have an integer common multiple because &radic;2 is irrational. There are 16 primitive Pythagorean triples with c &le; 100:



Generating Pythagorean triples
An effective way to generate Pythagorean triples is based on the observation that if m and n are two positive integers with m > n, then


 * $$a = m^2 - n^2,\,$$
 * $$b= 2mn,\,$$
 * $$c = m^2 + n^2,\,$$

is a Pythagorean triple. It is primitive if and only if m and n are coprime and one of them is even (if both n and m are odd, then a, b, and c will be even, and so the Pythagorean triple will not be primitive). Not every Pythagorean triple can be generated in this way, but every primitive triple (possibly after exchanging a and b) arises in this fashion from a unique pair of coprime numbers m > n. This shows that there are infinitely many primitive Pythagorean triples. This formula was given by Euclid (c. 300 B.C.) in his book Elements and is referred to as Euclid's formula.

An alternate form of the Euclid formula eliminates the negative sign by making use of the relation m = p + q and n = p:


 * $$a = q(2p + q) \,$$
 * $$b = 2p(p+q) \,$$
 * $$c = (p + q)^2 + p^2 \,$$

Also, it is easy to notice that the complex number $$m+in$$ when squared gives $$a+ib$$ as a result. Since $$|z^2|=|z|^2$$, $$\sqrt{a^2+b^2}=m^2+n^2=c$$ is an integer.

Properties of Pythagorean triples
The properties of primitive Pythagorean triples include:


 * Exactly one of a, b is odd; c is odd.
 * The area (A = ab/2) is an integer.
 * Exactly one of a, b is divisible by 3.
 * Exactly one of a, b is divisible by 4.
 * Exactly one of a, b, c is divisible by 5.
 * For any Pythagorean triple, ab is divisible by 12, and abc is divisible by 60.
 * Exactly one of a, b, (a + b), (a &minus; b) is divisible by 7.
 * At most one of a, b is a square.
 * Every integer greater than 2 is part of a Pythagorean triple.
 * There exist infinitely many Pythagorean triples whose hypotenuses are squares of natural numbers.
 * There exist infinitely many Pythagorean triples in which one of the legs is the square of a natural number.
 * For each natural number n, there exist n Pythagorean triples with different hypotenuses and the same area.
 * For each natural number n, there exist at least n different Pythagorean triples with the same leg a, where a is some natural number
 * For each natural number n, there exist at least n different triangles with the same hypotenuse.
 * In every Pythagorean triple, the radius of the incircle and the radii of the three excircles are natural numbers.
 * There is no Pythagorean triple in which the hypotenuse and one leg are the legs of another Pythagorean triple.
 * There is no Pythagorean triple in which the hypotenuse is equal to either 2a or 2b.

Some relationships
If $$ a^2 + b^2 = c^2 $$ is a primitive Pythagorean triple, where a is odd, then


 * $$\frac{c+a}{b}=\frac{m}{n},\,$$


 * $$ \frac{c+b+a}{c+b-a}= \frac{m}{n}\,$$


 * $$ b/(c-a)= m/n \,$$


 * $$(a+c-b)/(a+b-c)= m/n \,$$

where each fraction is reduced to lowest terms and m>n.

It can also be shown that


 * $$ b(m^2-n^2) = a(2mn)\,$$


 * $$ (m/n)b - a = c \,$$


 * $$ (n/m)b + a = c \,$$

Additional relationships among the sides:


 * $$ c - b = (m - n)^2 \,$$


 * $$ c + b = (m + n)^2 \,$$


 * $$ a^2 = c^2 - b^2 = (c - b)(c + b)\,$$


 * $$ c - a = (m^2 + n^2) - (m^2 - n^2) = 2n^2\,$$


 * $$ c = a + (m^2 + n^2) - (m^2 - n^2) = a + 2n^2\,$$


 * $$ a = c - (m^2 + n^2) - (m^2 - n^2) = c- 2n^2 \,$$



The radius, r, of the inscribed circle can be found by:


 * $$r = ab/(a+b+c) \,$$

where


 * $$a = 2r +1 \,  $$


 * $$b = 2r(r+1)\, $$


 * $$c = 2r^2 + 2r + 1\,$$


 * $$r = n (m-n) \, $$


 * $$ n/m = \frac{\sqrt{1+r^2}-1}{r} \ $$

The unknown sides of a triple can be calculated directly from the radius of the incircle, r, and the value of a single known side, a.
 * k = a &minus; 2r
 * b = 2r + (2 r2/k)
 * c = b+ k = 2r + (2r2 /k) + k

The perimeter P and area L of a primitive Pythagorean triple triangle are
 * P = a + b + c = 2m(m + n)
 * L = ab/2 = mn(m2 &minus; n2)

The shortest side will be a if one of the following conditions is met:
 * $$a < b  \,\!$$
 * $$m^2 - n^2 < 2mn \,\!$$
 * $$(m - n)^2 < 2n^2 \,\!$$
 * $$m - n < n \sqrt{2} \,\!$$
 * $$m < n (1 + \sqrt{2})\,\!$$

Unit circle relationships
An arbitrary rational slope, t on the unit circle can be written t = n/m where m and n are integers and m > n. Other unit circle relationships are shown below:


 * $$\cos\theta\ = {m^2-n^2 \over m^2+n^2} = {1-t^2 \over 1+t^2}= {a \over c}$$


 * $$\sin\theta\ = {2mn \over m^2+n^2} = {2t \over 1+t^2} = {b \over c}$$


 * $$\tan \theta\ = {2mn \over m^2-n^2} = {2t \over 1−t^2} = {b \over a}$$


 * $$x^2 + y^2 = 1 \,$$

Half-angle relationships

 * $$\tan\left({\theta \over 2}\right) = {n \over m},$$


 * $$\tan\left({\beta \over 2}\right) = {m-n \over m+n}.$$

A special case: the Platonic sequence
The case n = 1 of the above has been known for a long time. Proclus, in his commentary to the 47th Proposition of the first book of Euclid's Elements, describes it as follows:

''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.'' ... ''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 as that which was obtained by the other method.''

In equation form, this becomes:

a is odd:
 * side a = a :  side b =$$(a^2-1)/2 $$  :   side c=$$ (a^2 +1)/2 $$

a is even:
 * side a = a :  side b =$$(a/2)^2-1 $$  :   side c=$$ (a/2)^2 +1$$

In number theory, one modern mathematical generalization of this sequence is


 * $$(b, c) = \left(\frac{a}{2}\right)^2(1+a\bmod2)\mp\frac{1}{1+a\bmod2}.\,$$

That is to say, using modular arithmetic any natural number greater than 2 ("$$a$$" on the right hand side) can give us two complementary numbers ("$$b$$" and "$$c$$" on the left hand side) which together are a Pythagorean triple $$(a, b, c)$$.

Generalizations
A set of four positive integers a, b, c and d such that a2 + b2+ c2 = d2 is called a Pythagorean quadruple.

A generalization of the concept of Pythagorean triples is the search for triples of positive integers a, b, and c, such that an + bn = cn, for some n strictly greater than 2. Pierre de Fermat in 1637 claimed that no such triple exists, a claim that came to be known as Fermat's last theorem, though it was far from the last theorem Fermat discovered. The first proof was given by Andrew Wiles in 1994.

Other formulas for generating triples
I: Pythagoras (c. 540 B.C.) presented this formula for generating triples:


 * $$a =m \,:\, b = (m^2 - 1)/2 \,:\, c= (m^2 + 1)/2,$$

where m must be odd.

I(a): Declan O'Loughlin, Ireland, presented this formula for generating triples:


 * $$a =m \,:\, b = (m^2 - 4)/4 \,:\, c= (m^2 + b^2),$$

where m must be even.

II: Plato (c. 380 B. C.) is attributed with a second formula:


 * $$ a =2m \,:\, b= (m^2 - 1) \,:\, c=  (m^2 + 1), $$

where m is any natural number. Plato's formula will not directly produce triples whose longer side and hypotenuse differ by one, but can generate all primitive triples by dividing those triples that have a common factor of 2 by 2.

The methods below appear in various places, but without attribution as to their origin:

III. Given an integer n, the triple can be generated by the following two procedures:


 * $$ a= 2n + 1 \,:\,  b=2n(n + 1)  \,:\, c = 2n(n + 1) + 1 $$

Example: When n = 2 the triple produced is 5, 12, and 13 (This formula is actually the same as method I, substituting m with 2n + 1.)

IV. Given the integers n and x,


 * $$ a= 2x^2 + 2nx \,:\, b= 2nx + n^2 \,:\, c=2x^2 + 2nx + n^2 $$

Example: For n = 3 and x = 5, a = 80, b = 39, c = 89.

V. Triples can be calculated using this formula: $$2xy = z^2 $$, x,y,z > 0 where the following relations hold:

x = c &minus; b, y = c &minus; a, z = a + b &minus; c and a = x + z, b = y + z, c = x + y + z and r = z/2, where x, y, and z are the three sides of the triple and r is the radius of the inscribed circle.

Pythagorean triples can then be generated by choosing any even integer z.

x and y are any two factors of $$ z^2/2 $$.

Example: Choose z = 6. Then $$ z^2/2 =18. $$ The three factor-pairs of 18 are: (18, 1), (2, 9), and (6, 3). All three factor pairs will produce triples using the above equations.

z = 6, x = 18, y = 1 produces the triple a = 18 + 6 = 24, b = 1 + 6 = 7, c = 18 + 1 + 6 = 25.

z = 6, x = 2, y = 9 produces the triple a = 2 + 6 = 8, b = 9 + 6 = 15, c = 2 + 9 + 6 = 17.

z = 6, x = 6, y = 3 produces the triple a = 6 + 6 = 12, b = 3 + 6 = 9, c = 6 + 3 + 6 = 15.

VI. An infinity by infinity matrix M of Pythagorean triples (PNTs), which has some particularly desirable properties can be generated by taking:


 * $$a(r,k) = 4rk + 2k(k-1)\,$$


 * $$b(r, k) = 4r(r+k-1) - 2k + 1\,$$


 * $$c(r,k) = 4r(r+k-1) + 2k(k-1) + 1\,$$

where r is the row number and k is the column number. Note that a is always doubly even, while b and c are always odd. Not more than the first k rows in column k will have a > b. Each row is a family of PNTs with the hypotenuse c of each PNT in row r exceeding the even side a by the square of the rth odd number. The Pythagorean formula for generating PNTs (section I, above) with a and b reversed to make a the even side, and m being any natural number:


 * $$ b = 2m+1\,$$
 * $$ a = (b^2-1)/2\,$$
 * $$ c = a+1\,$$

yields the first row (r = 1) of M, and the Platonic formula (section II, above) using a = 4m instead of 2m, to eliminate derivative PNTs:


 * $$ a = 4m\,$$
 * $$ b = 4m^2-1\,$$
 * $$ c = b+2\,$$

yields the first column (k = 1) of M.

Each column is a family of PNTs with the hypotenuse of each PNT in column k exceeding the odd side b by twice the square of k. For example M(6,4) = {120, 209, 241} 241 &minus; 120 = 121, the square of 11, and 241 &minus; 209 = 32, which is twice the square of 4.

Below is a small portion of the matrix. The PNTs of row 1 are all relatively prime (primitive), but every other row contains derivative (not relatively prime) PNTs Iff the column number is a power of 2, the PNTs in that column are all primitive. For every odd prime factor p of the column number, the middle row of each group of p rows (r = (p+1)/2 + np, where n >= 0) will contain a PNT which is derivative. In the table below these are indicated by angle brackets. If j is 2 or a factor of k, then M(r, jk) is derivative iff M(r, k) is derivative. Fewer than 20% of the PNTs in M are derivative.

column-> 1               2                3                4                5 row  a    b    c      a    b    c      a    b    c      a    b    c      a    b    c  1    4    3    5     12    5   13     24    7   25     40    9   41     60   11   61 2   8   15   17     20   21   29       56   33   65     80   39   89  3   12   35   37     28   45   53     48   55   73     72   65   97     4   16   63   65     36   77   85     60   91  109     88  105  137    120  119  169  5   20   99  101     44  117  125      104  153  185    140  171  221  6   24  143  145     52  165  173     84  187  205    120  209  241    160  231  281

The a ' s of each column k are an arithmetic sequence with difference 4k, and the b ' s of each row r are an arithmetic sequence with difference 4r-2.

If the two legs of a PNT differ by 1, the longer leg and the hypotenuse form the coordinates of a larger PNT in M the legs of which differ by 1. M(1,1) = {4, 3, 5}. M(4,5) = {120, 119, 169}. M(120,169) = {137904, 137903, 195025}, etc. Thus, a Pythagorean triangle can be found, the acute angles of which are arbitrarily close to 45 degrees.

VII. Generalized Fibonacci Series: A pythagorean triple can be generated by using any two arbitrary integers, a and b using the following procedures:

a. select any two integers a and b

b. define c = a+b

c. define d = b+c

The integers a,b,c,d are a generalized Fibonacci series. The sides of the triple are computed as follows:

side 1 = $$ 2*b*c $$

side 2 = $$ a*d $$

hypotenuse = $$ b^2 + c^2 $$

example let a = 69 and b = 75, then c = 69+75 =144  and d= 75+144=219

side 1 = $$ 2*75*144=21600 $$

side 2 = $$ 69*219 = 15111 $$

hypotenuse = $$ 75^2 + 144^2 = 26361 $$

Proof: $$ sqr(21600^2  + 15111^2 ) = 26361 $$

http://www.mcs.surrey.ac.uk/Personal/R.Knott/Pythag/pythag.html

Parent Child Relationships
All primitive Pythagorean triples can be generated from the 3-4-5 triangle by using the 3 linear transformations below, where a,b,c are sides of a  triple:

column->              1                 2               3 new side a       new side b      new side c                             a - 2b + 2c       2a - b + 2c     2a - 2b + 3c a + 2b + 2c      2a + b + 2c     2a + 2b + 3c -a + 2b + 2c     -2a + b + 2c    -2a + 2b + 3c If one begins with 3,4,5 then all other primitive triples will eventually be produced. In other words, every primitive triple will be a “parent” to 3 additional primitive triples. example: Let a=3  b =4  c = 5

column-> new side a          new side b           new side c                           3-(2*4)+(2*5)=5     (2*3)-4+(2*5)=12   (2*3)-(2*4)+(3*5)=13 3+(2*4)+(2*5)=21   (2*3)+4+(2*5)=20   (2*3)+(2*4)+(3*5)=29                            -3+(2*4)+(2*5)=15   -(2*3)+4+(2*5)=8   -(2*3)+(2*4)+(3*5)=17

For further discussion of parent-child relationships in triples, see: http://mathworld.wolfram.com/PythagoreanTriple.html and “The Modular Tree of Pythagoras”, Robert Alperin, Department of Mathematics and Computer Science, San Jose State University, San Jose California) http://www.math.sjsu.edu/~alperin/pt.pdf and http://www.faust.fr.bw.schule.de/mhb/pythagen.htm