Pythagorean triple

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The Pythagorean theorem: a2 + b2 = c2
The Pythagorean theorem: a2 + b2 = c2

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 = √2 is right, but (1, 1, √2) is not a Pythagorean triple because √2 is not an integer. Moreover, 1 and √2 do not have an integer common multiple because √2 is irrational. There are 16 primitive Pythagorean triples with c ≤ 100:

( 3, 4, 5) ( 5, 12, 13) ( 7, 24, 25) ( 8, 15, 17)
( 9, 40, 41) (11, 60, 61) (12, 35, 37) (13, 84, 85)
(16, 63, 65) (20, 21, 29) (28, 45, 53) (33, 56, 65)
(36, 77, 85) (39, 80, 89) (48, 55, 73) (65, 72, 97)
A scatter plot of the first Pythagorean triples within 4500 covering only integral values.
A scatter plot of the first Pythagorean triples within 4500 covering only integral values.

Contents

[edit] Generating a triple

Another Triple is 9, 12, 15

This classic formula was given by Euclid (c. 300 B.C.) in his book Elements and is often referred to as Euclid's formula.

a = (m^2 - n^2) \,:\, b = (2mn) \,:\, c = (m^2 + n^2)

where m and n are two positive integers with m > n. The triple generated will be primitive if and only if m and n are coprime and exactly 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 all non-primitive Pythagorean triples can be generated with this formula, 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 (combine, for example, 2 with any of the infinitely many other prime numbers). Relationships to the integers m,n from this equation are referenced throughout the rest of this article.

The relation between (m,n) and (a,b) is that in the complex plane a + ib is m + in squared.

An alternative 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.

See also: Other formulas for generating triples below for additional equations that can be used to generate triples.

[edit] Properties of primitive 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.
  • Exactly one of a, b, (a + b), (b − a) is divisible by 7.
  • All prime factors of c are primes of the form 4n+1.
  • At most one of a, b is a square.
  • Every integer greater than 2 that is not congruent to 2 mod 4 is part of a primitive Pythagorean triple. Examples of integers not part of a primitive pythagorean triple: 6,10,14,18
  • Every integer greater than 2 is part of a primitive or non-primitive Pythagorean triple, for example, the integers 6,10,14, and 18 are not part of primitive triples, but are part of the non-primitive triples 6,8,10; 14,48,50 and 18,80,82.
  • 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.
  • There exist infinitely many Pythagorean triples in which the hypotenuse and the longer of the two legs differ by exactly one.
  • There exist infinitely many Pythagorean triples in which the hypotenuse and the longer of the two legs differ by exactly two.
  • There are no primitive Pythagorean triples in which the hypotenuse and a leg differ by a prime number greater than 2.
  • 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. (Actually the radius of the incircle can be shown to be r = n(m − n) )
  • There is no Pythagorean triple in which the hypotenuse and one leg are the legs of another Pythagorean triple.
  • In a pythagorean triplet a+b=c+2[(c-a)(c-b)/2]^1/2.
  • (c-a)(c-b)/2 is always a perfect square.

[edit] Some relationships

If a2 + b2 = c2 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)= \frac{m}{n} \,
(a+c-b)/(a+b-c)= \frac{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 \,
Right Triangle with inscribed circle of radius r
Right Triangle with inscribed circle of radius r

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

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

for primitive triples:

r = n(m-n) \,

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 − 2r
b = 2r + (2 r2/k)
c = b+ k = 2r + (2r2 /k) + k

The solution to the 'Incircles' problem shows that, for any circle whose radius is a whole number k, we are guaranteed at least one right angled triangle containing this circle as its inscribed circle where the lengths of the sides of the triangle are a primitive Pythagorean triple:

a=2k(k+1)
b=2k+1
c=2k2+2k+1

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 − 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})\,\!
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 \,

More relationships among the sides:

c^4=(a^2-b^2)^2+(2ab)^2\,
a^4=(c^2+b^2)^2-(2cb)^2\,
b^4=(c^2+a^2)^2-(2ca)^2\,
a^2b^2=(c^2+ab)^2-(ca+cb)^2=(c^2-ab)^2-(ca-cb)^2\,

see: http://www.geocities.com/fredlb37/node8.html

If two numbers of a triple are known, the third can be found using the Pythagorean theorem.

[edit] 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:

Right triangle with slope, t
Right triangle with slope, t
\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 \,

[edit] Half-angle relationships

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

[edit] A special case: the Platonic sequence

The case n = 1 of the more general construction of Pythagorean triples 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 that which was obtained by the other method.

In equation form, this becomes:

a is odd (Pythagoras, c. 540 BC):

\mbox{side }a : \mbox{side }b = {a^2 - 1 \over 2} : \mbox{side }c = {a^2 + 1 \over 2}.

a is even (Plato, c. 380 BC):

\mbox{side }a : \mbox{side }b = \left({a \over 2}\right)^2 - 1 : \mbox{side }c = \left({a \over 2}\right)^2 + 1

It can be shown that all Pythagorean triples are derivatives of the basic Platonic sequence (x,y,z) = p, (p2 - 1)/2 and (p2 + 1)/2 by allowing a to take non-integer rational values. If p is replaced with the rational fraction m/n in the sequence, the 'standard' triple generator 2mn, m2 - n2  and m2 + n2  results. It follows that every triple has a corresponding rational p value which can be used to generate a similar (i.e. equiangular) triangle with rational sides in the same proportion as the original. For example, the Platonic equivalent of (6,8,10) is (3/2; 2, 5/2). The Platonic sequence itself can be derived by following the steps for 'splitting the square' described in Diophantus II.VIII.

[edit] Other formulas for generating triples

I, II: Pythagoras' and Plato's formulas have been described above. The methods below appear in various sources, often 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.)

Alternatively, one can generate triples from even integers using the following formulas:

a= 2m \,:\, b=m^2 - 1 \,:\, c = m^2 + 1

Example: When m = 4 the triple produced is 8, 15, and 17 (This formula is another specific case of method I, substituting n with 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. (This formula is actually the same as method I, substituting m and n with n+x and x.)

V. Triples can be calculated using this formula: 2xy = z2, x,y,z > 0 where the following relations hold:

x = c − b, y = c − a, z = a + b − 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 z2 / 2.

Example: Choose z = 6. Then z2 / 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:

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

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:

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

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 − 120 = 121, the square of the sixth odd number (11), and 241 − 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 if and only if 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    <36   27   45>    56   33   65     80   39   89
 3   12   35   37     28   45   53     48   55   73     72   65   97   <100   75  125>
 4   16   63   65     36   77   85     60   91  109     88  105  137    120  119  169
 5   20   99  101     44  117  125    <72  135  153>   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. The a's, b's, and c's of any row or column are each monotonically increasing.

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. As Martin (1875) describes, each such triple has the form

(2P_{n}P_{n+1}, P_{n+1}^2 - P_{n}^2, P_{n+1}^2 + P_{n}^2).

where Pi are the Pell numbers.

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 = 2bc

side 2 = ad

hypotenuse = b2 + c2

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

side 1 = 2\cdot 75\cdot 144=21600

side 2 = 69\cdot 219 = 15111

hypotenuse = 752 + 1442 = 26361

216002 + 151112 = 263612

VIII. Progression of Whole and Fractional Numbers:

Take a progression of whole and fractional numbers: 1 1/3, 2 2/5 , 3 3/7 , 4 4/9 etc. The properties of this progression are: a) the whole numbers are those of the common series and have unity as their common difference b) the numerators of the fractions, annexed to the whole numbers, are also the natural numbers. 3) the denominators of the fractions are the odd numbers, 3,5,7, etc.

To calculate a pythagorean triple:

select any term of this progression and reduce it to an improper fraction. For example, take the term 3 3/7. The improper fraction is 24/7. The numbers 7 and 24 are the sides, a and b, of a right triangle. The hypotenuse is one greater than the largest side.

1 1/3 yields the 3,4,5 triple; 2 2/5 gives 5,12,13 ; 3 3/7 yields gives 7,24, 25 ; 4 4/9 gives 9,40,41 and so forth.

IX. Generating Triples using a Square:

Start with any square number n. Express that number in the form x(x + 2y), then y2 will produce another square such that n + y2 = z2. For instance:

let n = 9, 1(1 + 8) = 9, (8 / 2)2 = 16, and 9 + 16 = 25.

let n = 36, 2(2 + 16) = 36, (16 / 2)2 = 64, and 36 + 64 = 100.

This works because x(x + 2y) = x2 + 2xy. If we add y2, our expression becomes x2 + 2xy + y2, which factors into the form (x + y)2

X. Generating Triples When One Side is Known:

Start with any integer b. Use this relation from the Euclid formula: b = 2mn. If b is odd, then multiply b by 2. Identify all factor-pairs (m,n) of b and use the Euclid equations to calculate the remaining sides of the triple.

Examples: Let b =24 (e.g. the known side is even)

24 = 2mn so that 12 = mn . The factor pairs (m,n) of 12 are (12,1), (6,2) and (4,3). The three triples are therefore:

a = (m^2-n^2) \,:\ b =2mn \,:\ c = (m^2 + n^2)

a=12^2-1^2 =143 \,:\ b = 24 \,:\ c = (m^2 + n^2)=145

a= 6^2-2^2 =32 \,:\ b = 24 \,:\ c = (6^2 + 2^2)=40

a = 4^2-3^2 =7 \,:\ b = 24 \,:\ c = (4^2 + 3^2)=25

Let b =35 (e.g. the known side is odd)

The two unknown sides could also be calculated by making use of the relation a = (m2 − n2) . This would be a factoring exercise in finding the difference of two squares, but a simpler approach is to multiply the known side by two and continue as before :

70 = 2mn so that 35 = mn . The factor pairs (m,n) of 35 are (35,1), (7,5).

The two triples are therefore (note that is necessary to remove the factor of 2 which was introduced):

a=(35^2-1^2)/2=612 \,:\ b = 70/2=35 \,:\ c = (35^2 + 1^2)/2=613

a=(7^2-5^2)/2=12 \,:\ b = 70/2=35 \,:\ c = (7^2 + 5^2)/2=37

[edit] Parent/child relationships

All primitive Pythagorean triples can be generated from the 3-4-5 triangle by using the 3 linear transformations T1, T2, T3 below, where a ,b, c are sides of a triple:

                   new side a        new side b      new side c     
          T1:     a - 2b + 2c       2a - b + 2c     2a - 2b + 3c 
          T2:     a + 2b + 2c       2a + b + 2c     2a + 2b + 3c
          T3:    -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.

            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

The linear transformations T1, T2, and T3 have a geometric interpretation in the language of quadratic forms. They are closely related to (but are not equal to) reflections generating the orthogonal group of x2 + y2 - z2 over the integers.

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

[edit] Generalizations

There are several ways to generalize the concept of Pythagorean triples.

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 because it took longer than any other theorem by Fermat to be proven or disproven. The first proof was given by Andrew Wiles in 1994.

Another generalization is searching for sets of n+1 positive integers for which the nth power of the last is the sum of the nth powers of the previous terms. The smallest sets for known values of n are:

  • n=3: {3, 4, 5, 6}.
  • n=4: {30, 120, 272, 315, 353}
  • n=5: {19, 43, 46, 47, 67, 72}
  • n=7: {127, 258, 266, 413, 430, 439, 525, 568}
  • n=8: {90, 223, 478, 524, 748, 1088, 1190, 1324, 1409}

[edit] See also

[edit] References

[edit] External links