Pythagorean triple

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, stating that every right triangle has side lengths satisfying the formula a² + b² = c²; thus, Pythagorean triples describe the three integer side lengths of a right triangle. However, right triangles with non-integer sides do not form Pythagorean triples. 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 legs (a,b) of the first Pythagorean triples with a and b less than 4500.

Generating a triple

A plot of triples generated by Euclid's formula map out part of the z² = x² + y² cone. A constant m or n trace out part of a parabola on the cone.

Euclid's formula is a fundamental formula for generating Pythagorean triples given an arbitrary pair of positive integers m and n with m > n. The formula states that the integers

$a=2mn\,:\,b=m^{2}-n^{2}\,:\,c=m^{2}+n^{2}$

form a Pythagorean triple. The triple generated by Euclid's formula is 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 triple will not be primitive. Every primitive triple (possibly after exchanging a and b) arises from a unique pair of coprime numbers m, n, one of which is even. It follows that there are infinitely many primitive Pythagorean triples. This relationship of a and b to m and n from Euclid's formula is referenced throughout the rest of this article.

Despite generating all primitive triples, Euclid's formula does not produce all triples. This can be remedied by inserting an additional parameter k to the formula. The following will generate all Pythagorean triples (although not uniquely):

$a=k\cdot (2mn)\,:\,b=k\cdot (m^{2}-n^{2})\,:\,c=k\cdot (m^{2}+n^{2})$

where m, n, and k are positive integers with m > n.

An alternative form of Euclid's formula introduces a different pair of positive integer parameters p and q to eliminate the need for imposing further conditions on m and n. Letting m = p + 2q - 1 and n = p, then m - n = 2q - 1 is necessarily odd, and Euclid's formula becomes:

$a=2p(p+2q-1)\,:\,b=(2q-1)(2p+2q-1)\,:\,c=(p+2q-1)^{2}+p^{2}.$

Many other formulas for generating triples have been developed since the time of Euclid.

Elementary properties of primitive Pythagorean triples

The properties of primitive Pythagorean triples include:

• 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. This is particularly useful in checking if a given triplet of numbers is a pythagorean triple, but it is only a necessary condition, not a sufficient one. The triple {6, 12, 18} passes (c − a)(c − b)/2 but is not a PNT. A simpler, more powerful test is, (by naming the even leg a): (c − a) and (c − b)/2 are both perfect squares. This is both necessary and sufficient for the triple to be a PNT, but the PNT may be derivative. If any two sides of a PNT are relatively prime, it is a primitive PNT.
• 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 primitive Pythagorean triples whose hypotenuses are squares of natural numbers.
• There exist infinitely many primitive 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 (such triples are necessarily primitive). Generalization: For every odd integer j, there exist infinitely many primitive Pythagorean triples in which the hypoteneuse and the even leg differ by j ².
• There exist infinitely many primitive Pythagorean triples in which the hypotenuse and the longer of the two legs differ by exactly two. Generalization: For every integer k > 0, there exist infinitely many primitive Pythagorean triples in which the hypotenuse and the odd leg differ by $2k^{2}$.
• If j and k are odd positive integers, not necessarily unequal, there is exactly one primitive Pythagorean triple with $a+j^{2}=c=b+2^{k}.$
• The hypoteneuse of every primitive Pythagorean triangle exceeds the even leg by the square of an odd integer j, and exceeds the odd leg by twice the square of an integer k > 0, from which it follows that:
• 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 are no Pythagorean triplets in which the hypotenuse and one leg are the legs of another Pythagorean triple.

Some relationships

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 the right triangle corresponding to a primitive Pythagorean triple triangle are

P = a + b + c = 2m(m + n)

L = ab/2 = mn(m² − n²)

Additional Relationships:

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

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

$b/(c-a)={\frac {m}{n}}\,$

$(a+c-b)/(a+b-c)={\frac {m}{n}}\,$

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

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

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

$x^{2}+y^{2}=1\,$

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

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

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.

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.

Other formulae for generating triples

Besides Euclid's formula, many other formulas which only generate Pythagorean triples have been developed.

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: $2xy=z^{2}$, 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 $z^{2}/2$.

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.

Euclid's well-known formula (see Generating a triple above):

$a=2mn\,:\,b=m^{2}-n^{2}\,:\,c=m^{2}+n^{2}$

can generate an infinity by infinity matrix E of Pythagorean triples (PNTs) by making m the row number and n the column number (or vice versa). E will contain every mostly derivative (non-relatively prime) PNTs because a PNT in E will not be relatively prime (primitive) if m and n are both even, or both odd, or if they have a common factor > 1. Also, E(n,m) will be the same PNT as E(m,n), so nearly half the PNTs in E will be duplicates(but with b negative) and more than half will be derivative. And if m = n (the main diagonal), the result will be a degenerate PNT, {c, 0, c}. Below is a portion of E, with the primitive PNTs enclosed in ::s.

 2,  0,  2    ::4,  3,  5::    6,  8  10    ::8, 15, 17::   10, 24, 26    ::12, 35, 37::
 4, -3,  5      8,  0,  8   ::12,  5  13::   16, 12, 20   ::20, 21, 29::    24, 32, 40   
 6, -8, 10     12, -5, 13     18,  0  18   ::24,  7, 25::   30, 16, 34     <36, 27, 45>
 8,-15, 17     16,-12, 20     24, -7  25     32,  0, 32   ::40,  9, 41<::   48, 20, 52
10,-24, 26     20,-21, 29     30,-16  34     40, -9, 41     50,  0, 50    ::60, 11, 61::
12,-35, 37     24,-32, 40     36,-27  45     48,-20, 52     60,-11, 61      72,  0, 72

Another infinity by infinity matrix M of PNTs, which has some particularly desirable properties (no duplicates, fewer derivatives, corresponding entries (a, b, c) in each row and each column increase monotonically, and each row and each column of M is a different single family of PNTs--see below)) 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. These formulae for a, b, and c can be verified by substituting the right-hand side of each of the above equations into

$a^{2}+b^{2}=c^{2}\,$

$[4rk+2k(k-1)]^{2}+[4r(r+k-1)-2k+1]^{2}=[4r(r+k-1)+2k(k-1)+1]^{2}\,$

Squaring out the above expressions (and inserting periods in the first two to make like terms line up) gives us:

$a^{2}=4k^{4}-8k^{3}+4k^{2}.....................+.....................16rk^{3}+16r^{2}k^{2}...........-...........16rk^{2}\,$

$b^{2}=................4k^{2}-4k+16r^{4}-32r^{3}+24r^{2}-8r......+......16r^{2}k^{2}+32r^{3}k-48r^{2}k-16rk^{2}+24rk+1\,$

.............._________________________________________________________________________________________________

$c^{2}=4k^{4}-8k^{3}+8k^{2}-4k+16r^{4}-32r^{3}+24r^{2}-8r+16rk^{3}+32r^{2}k^{2}+32r^{3}k-48r^{2}k-32rk^{2}+24rk+1\,$

Note that always, a is doubly even, b and c are odd.

A one-to-one mapping can be established between all of M and those PNTs of E for which one of {m,n} is even, the other is odd, and m > n:

$m=2r+k-1\,:\,n=k\,$

$r=(m+1-n)\,:\,k=n\,$

therefore all primitive PNTs are in M, albeit many derivative PNTs in E are not in M.

Most of the PNTs in M will have a < b; a > b will occur only when r < (k√2+1)/2

Since (k√2+1) is never an integer, therefore r = (k√2+1)/2 can't occur; if r > (k√2+1)/2, which it is for all but a finite number (< k, if k > 2) of PNTs in column k, 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 r^th odd number.

$c(r,k)-a(r,k)=[4r(r+k-1)+2k(k-1)+1]-[4rk+2k(k-1)]\,$

$c(r,k)-a(r,k)=[4r(r+k-1)+1]-[4rk]\,$

$c(r,k)-a(r,k)=[4r^{2}+4rk-4r+1]-[4rk]\,$

$c(r,k)-a(r,k)=[4r^{2}-4r+1]=(2r-1)^{2}\,$

$c(r,k)-a(r,k)=(2r-1)^{2}\,$ = the square of the r^th 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, twice the square of 4.

$c(r,k)-b(r,k)=[4r(r+k-1)+2k(k-1)+1]-[4r(r+k-1)-2k+1]\,$

$c(r,k)-b(r,k)=[+2k(k-1)]-[-2k]\,$

$c(r,k)-b(r,k)=[2k^{2}-2k]-[-2k]=2k^{2}\,$

$c(r,k)-b(r,k)=2k^{2}\,$

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
 7   28  195  197     60  221  229     96  247  265    136  273  305    180  299  349
 8   32  255  257     68  285  293   <108  315  333>   152  355  373   <200  375  425>

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:

$a(r+1,k)-a(r,k)=[4(r+1)k+2k(k-1)]-[4rk+2k(k-1)]=[4k(r+1-r)]=4k>0\,$

$a(r,k+1)-a(r,k)=[4r(k+1)+2(k+1)k]-[4rk+2k(k-1)]=4(r+k)>0\,$

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

$b(r+1,k)-b(r,k)=[4(r+1)(r+1+k-1)]-[4r(r+k-1)]=4(2r+k)>0\,$

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

$b(r,k+1)-b(r,k)=[4r(r+k)-2k-2]-[4r(r+k-1)-2k]=(4r-2)>0\,$

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

$c(r+1,k)-c(r,k)=[4(r+1)(r+k)]-[4r(r+k-1)]=4[(r^{2}+rk+r+k)-(r^{2}+rk-r)=4(2r+k)>0\,$

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

$c(r,k+1)-c(r,k)=[4r(r+k)+2(k+1)k)]-[4r(r+k-1)+2k(k-1)]=4r+2k[(k+1)-(k-1)]=4(r+k)>0\,$

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(21,29) = {4060, 4059, 5741}, M(120,169) = {137904, 137903, 195025}, etc. Note that this gives only PNTs where a > b, but Sierpinski (2003) gives a formula for finding near-isosceles PNTs which alternately gives a = b+1 and b = a+1. Thus, a Pythagorean triangle can be found, the acute angles of which are arbitrarily close (but never equal) 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 $P_{i}$ 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 = $b^{2}+c^{2}$

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 = $75^{2}+144^{2}=26361$

$21600^{2}+15111^{2}=26361^{2}$

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 $y^{2}$ will produce another square such that $n+y^{2}=z^{2}$. 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 $x(x+2y)=x^{2}+2xy$. If we add $y^{2}$, our expression becomes $x^{2}+2xy+y^{2}$, which factors into the form $(x+y)^{2}$.

Source: "Sequence A129861 in The On-Line Encyclopedia of Integer Sequences". Online Encyclopedia of Integer Sequences. N. J. A. Sloane. May 23 2007. Retrieved 2008-10-18. {{cite web}}: Check date values in: |date= (help)

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 $a=(m^{2}-n^{2})$ . 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):

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

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

XI.

Generating Triples Using Quadratic Equations:

There are several methods for defining quadratic equations for calculating each leg of a Pythagorean triple. A simple method is to modify the standard Euclid equation by adding a variable “x” to each m and n pair. The “m,n” pair is treated as a constant while the value of x is varied to produce a “family” of triples based on the selected triple. An arbitrary coefficient can be placed in front of the “x” value on either m or n, which has the effect causing the resulting equation to systematically “skip” through the triples. For example, let’s use the triple 20,21, 29, when can be calculated from the Euclid equations with a value of m=5 and n=2. Also, let’s arbitrarily put the coefficient of 4 in front of the “x” in the “m” term.

Let m1 = (4x + m) and n1= (x + n)

Hence, substituting the values of m and n:

$SideA=2*(m1*n1)=2*(4x+5)(x+2)=8x^{2}+26x+20$

$SideB=(m1)^{2}-(n1)^{2}=(4x+5)^{2}-(x+2)^{2}=15x^{2}+36x+21$

$SideC=(m1)^{2}+(n1)^{2}=(4x+5)^{2}+(x+2)^{2}=17x^{2}+44x+29$

Note that the original triple comprises the constant term in each of the respective quadratic equations. Below is a sample output from these equations. Note that the effect of these equations is to cause the “m” value in the Euclid equations to increment in steps of 4, while the “n” value increments by 1.

               x     side a       side b      side c     m  n    
               0       20          21            29      5  2 
               1       54          72            90      9  3
               2      104         153           185      13 4
               3      170         264           314      17 5
               4      252         405           477      21 6

Geometry of Euclid's formula

The rational points on a circle correspond, under stereographic projection, to the rational points of the line.

Euclid's formulas for a Pythagorean triple

$a=2mn,\quad b=m^{2}-n^{2},\quad c=m^{2}+n^{2}$

can be understood in terms of the geometry of rational number points on the unit circle (Trautman 1998). To motivate this, consider a right triangle with legs a and b, and hypotenuse c, where a, b, and c are positive integers. By the Pythagorean theorem, a² + b^{2 = c2} or, dividing both sides by c²,

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

Geometrically, the point in the Cartesian plane with coordinates

$x={\frac {a}{c}},\quad y={\frac {b}{c}}$

is on the unit circle x² + y² = 1. In this equation, the coordinates x and y are given by rational numbers. Conversely, any point on the unit circle whose coordinates x, y are rational numbers gives rise to a primitive Pythagorean triple. Indeed, write x and y as fractions in lowest terms:

$x={\frac {a}{c}},\quad y={\frac {b}{c}}$

where the greatest common divisor of a, b, and c is 1. Then, since x and y are on the unit circle,

$\left({\frac {a}{c}}\right)^{2}+\left({\frac {b}{c}}\right)^{2}=1\implies a^{2}+b^{2}=c^{2},$

as claimed.

Stereographic projection of the unit circle onto the x-axis. Given a point P on the unit circle, draw a line from P to the point N = (0,1) (the north pole). The point P′ where the line intersects the x-axis is the stereographic projection of P. Inversely, starting with a point P′ on the x-axis, and drawing a line from P′ to N, the inverse stereographic projection is the point P where the line intersects the unit circle.

There is therefore a correspondence between points on the unit circle with rational coordinates and primitive Pythagorean triples. At this point, Euler's formulas can be derived either by methods of trigonometry or equivalently by using the stereographic projection.

For this, suppose that P′ is a point on the x-axis with rational coordinates P′(m/n,0). Then, it can be shown by basic algebra that the point P has coordinates

$\left({\frac {2\left({\frac {m}{n}}\right)}{\left({\frac {m}{n}}\right)^{2}+1}},{\frac {\left({\frac {m}{n}}\right)^{2}-1}{\left({\frac {m}{n}}\right)^{2}+1}}\right)=\left({\frac {2mn}{m^{2}+n^{2}}},{\frac {m^{2}-n^{2}}{m^{2}+n^{2}}}\right).$

This establishes that each rational point of the x-axis goes over to a rational point of the unit circle. The converse, that every rational point of the unit circle comes from such a point of the x-axis, follows by applying the inverse stereographic projection. Suppose that P(x, y) is a point of the unit circle with x and y rational numbers. Then the point P′ obtained by stereographic projection onto the x-axis has coordinates

$\left({\frac {x}{1-y}},0\right)$

which is rational.

In terms of algebraic geometry, the algebraic variety of rational points on the unit circle is birational to the affine line over the rational numbers. The unit circle is thus called a rational curve, and it is this fact which enables an explicit parameterization of the (rational number) points on it by means of rational functions.

Spinors and the modular group

Pythagorean triples can likewise be encoded into a matrix of the form

$X={\begin{bmatrix}c+b&a\\a&c-b\end{bmatrix}}.$

A matrix of this form is symmetric. Furthermore, the determinant of X is

$\det X=c^{2}-a^{2}-b^{2}\,$

which is zero precisely when (a,b,c) is a Pythagorean triple. If X corresponds to a Pythagorean triple, then as a matrix it must have rank 1. Since X is symmetric, it follow from a result in linear algebra that there is a vector ξ = [m n]^T such that the outer product

$X=2{\begin{bmatrix}m\\n\end{bmatrix}}[m\ n]=2\xi \xi ^{T}\,$ (1)

holds, where the T denotes the matrix transpose. The vector ξ is called a spinor (for the Lorentz group SO(1,2)). In abstract terms, the Euler formula means that each primitive Pythagorean triple can be written as the outer product with itself of a spinor with integer entries, as in (1).

The modular group Γ is the set of 2×2 matrices with integer entries

$A={\begin{bmatrix}\alpha &\beta \\\gamma &\delta \end{bmatrix}}$

with determinant equal to one: αδ − βγ = 1. This set forms a group, since the inverse of a matrix in Γ is again in Γ, as is the product of two matrices in Γ. The modular group acts on the collection of all integer spinors. Furthermore, the group is transitive on the collection of integer spinors with relatively prime entries. For if [m n]^T has relatively prime entries, then

${\begin{bmatrix}m&-v\\n&u\end{bmatrix}}{\begin{bmatrix}1\\0\end{bmatrix}}={\begin{bmatrix}m\\n\end{bmatrix}}$

where u and v are selected (by the Euclidean algorithm) so that mu + nv = 1.

By acting on the spinor ξ in (1), the action of Γ goes over to an action on Pythagorean triples, provided one allows for triples with possibly negative components. Thus if A is a matrix in Γ, then

$2(A\xi )(A\xi )^{T}=AXA^{T}\,$ (2)

gives rise to an action on the matrix X in (1). This does not give a well-defined action on primitive triples, since it may take a primitive triple to an imprimitive one. It is convenient at this point (per Trautman 1998) to call a triple (a,b,c) standard if c > 0 and either (a,b,c) are relatively prime or (a/2,b/2,c/2) are relatively prime with a/2 odd. If the spinor [m n]^T has relatively prime entries, then the associated triple (a,b,c) determined by (1) is a standard triple. It follows that the action of the modular group is transitive on the set of standard triples.

Alternatively, restrict attention to those values of m and n for which m is odd and n is even. Let the subgroup Γ(2) of Γ be the kernel of the group homomorphism

$\Gamma =SL(2,\mathbf {Z} )\to SL(2,\mathbf {Z} _{2})$

where SL(2,Z₂) is the special linear group over the finite field Z₂ of integers modulo 2. Then Γ(2) is the group of unimodular transformations which preserve the parity of each entry. Thus if the first entry of ξ is odd and the second entry is even, then the same is true of Aξ for all A ∈ Γ(2). In fact, under the action (2), the group Γ(2) acts transitively on the collection of primitive Pythagorean triples (Alperin 2005).

The group Γ(2) is the free group whose generators are the matrices

$U={\begin{bmatrix}1&2\\0&1\end{bmatrix}},\qquad L={\begin{bmatrix}1&0\\2&1\end{bmatrix}}.$

Consequently, every primitive Pythagorean triple can be obtained in a unique way as a product of copies of the matrices U and L.

Parent/child relationships

By a result of Barning (1963), 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 (Alperin 2005).

Distribution of triples

A scatter plot of the legs (a,b) of the first Pythagorean triples with a and b less than 4500.

There are a number of results on the distribution of Pythagorean triples. In the scatter plot, a number of obvious patterns are already apparent. Whenever the legs (a,b) of a primitive triple appears in the plot, the all integer multiples of (a,b) must also appear in the plot, and this property produces the appearance of lines in the diagram.

Within the scatter, there are sets of parabolic patterns of high density of points all with their foci at the origin, opening up in all four directions. The points cluster along the parabolic arc in noisy manner as they do in Euler's totient function. Different parabolas intersect at the axes and appear as if to reflect off the axis with an incidence angle of 45 degrees, with a third parabola coming in in a perpendicular fashion. Within this quadrant, each arc centered around origin shows that section of the parabola that between its tip and its intersection with its semi-latus rectum. As mentioned earlier, the pattern repeats as you move away from or towards the origin. Albert Fässler, and others, provides insights into the significance of these real arcs and additional scatter plots that assist in recognizing those pattern in the diagram in this article.

Generalizations

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

Pythagorean quadruple

A set of four positive integers a, b, c and d such that a² + b²+ c² = d² is called a Pythagorean quadruple. The simplest example is (1,2,2,3), since 1² + 2² + 2² = 3². The next simplest (primitive) example is (2,3,6,7), since 2² + 3² + 6² = 7².

Fermat's Last Theorem

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 conjecture by Fermat to be proven or disproven. The first proof was given by Andrew Wiles in 1994.

n nth powers summing to an nth power

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}

A slightly different generalization allows the sum of (k+1) nth powers to equal the sum of (n-k) nth powers. For example:

• (n=3): 1³ + 12³ = 9³ + 10³, made famous by Hardy's recollection of a conversation with Ramanujan about the number 1729 being the smallest number that can be expressed as a sum of two cubes in two distinct ways.

Integral triangle triples

An integral triangle is one with distinct positive integer sides whose area is also an integer. The lengths of the sides of such a triangle form an integral triangle triple (or ITT) (a, b, c) provided a < b < c. Clearly, any PT is an ITT, since in a PT at least one of the legs a, b must be even, so that the area ab/2 is an integer. Not every ITT is a PT, however, as the example (4, 13, 15) with area 24 shows.

If (a, b, c) is an ITT, so is (ma, mb, mc) where m is any positive integer greater than one. The ITT (a, b, c) is primitive provided a, b, c are relatively prime (as with a PT). Here are a few of the simplest primitive ITTs which are not PTs:

(4, 13, 15) with area 24

(3, 25, 26) with area 36

(7, 15, 20) with area 42

(6, 25, 29) with area 60

(11, 13, 20) with area 66

(13, 14, 15) with area 84

(13, 20, 21) with area 126

By Heron's formula, the extra condition on a triple of positive integers (a, b, c) with a < b < c is that

(a² + b² + c²)² - 2(a⁴ + b⁴ + c⁴)

or equivalently

2(a²b² + a²c² + b²c²) - (a⁴ + b⁴ + c⁴)

be a nonzero perfect square divisible by 16.

See also

• Heronian triangle
• Pythagorean prime
• Nonhypotenuse number
• Modular arithmetic
• Trigonometric identity
• Tangent half-angle formula
• Plimpton 322
• Diophantus II.VIII
• Hilbert's theorem 90

References

• Alperin, Roger C. (2005), "The modular tree of Pythagoras" (PDF), American Mathematics Monthly, 112: 807–816, MR2179860
• Barning, F.J.M. (1963), "On Pythagorean and quasi-Pythagorean triangles and a generation process with the help of unimodular matrices", Math. Centrum Amsterdam Afd. Zuivere Wisk., ZW-011: 37 Template:Nl
• Eckert, Ernest (1992), "Primitive Pythagorean triples", The College Mathematics Journal, 23 (5): 413–417
• Elkies, Noam, Pythagorean triples and Hilbert's theorem 90 (PDF).
• Heath, Thomas (1956), The Thirteen Books of Euclid's Elements Vol. 1 (Books I and II) (2nd ed.), Dover Publications, ISBN 0-486-60088-2.
• Teigen, M. G.; Hadwin, D. W. (1971), "On Generating Pythagorean Triples", The American Mathematical Monthly, 78 (4): 378–379
• Martin, Artemas (1875), "Rational right angled triangles nearly isosceles", The Analyst, 3 (2): 47–50, doi:10.2307/2635906
• McCullough, Darryl (2005), "Height and excess of Pythagorean triples" (PDF), Mathematics Magazine, 78 (1)
• Romik, Dan (2004), The dynamics of Pythagorean triples, arXiv:math.DS/0406512
• Sierpinski, Wacław (2003), Pythagorean Triangles, Dover Publications, ISBN 0-486-43278-5.
• Trautman, Andrzej (1998), "Pythagorean spinors and Penrose twistors", Geometric universe {{citation}}: Unknown parameter |editors= ignored (|editor= suggested) (help).

External links

• http://mathworld.wolfram.com/PythagoreanTriple.html has an extensive discussion of Pythagorean triples.
• Pythagorean Triples at cut-the-knot Interactive Applet showing unit circle relationships to Pythagorean Triples
• The Trinary Tree(s) underlying Primitive Pythagorean Triples at cut-the-knot
• http://www.math.rutgers.edu/~erowland/pythagoreantriples.html Theoretical properties of the Pythagorean Triples and connections to geometry
• Clifford Algebras and Euclid's Parameterization of Pythagorean triples
• Pythagorean Triplets
• http://www.mcs.surrey.ac.uk/Personal/R.Knott/Pythag/pythag.html Discussion of Properties of Pythagorean triples, Interactive Calculators, Puzzles and Problems
• http://people.wcsu.edu/sandifere/Academics/2007Spring/Mat342/PythagTrip02.pdf Generating Pythagorean Triples Using Arithmatic Progressions
• Parameterization of Pythagorean Triples by a single triple of polynomials.
• Curious Consequences of a Miscopied Quadratic
• Solutions to Quadratic Compatible Pairs in relation to Pythagorean Triples
• The negative Pell equation and Pythagorean triples
• The Remarkable Incircle of a Triangle
• Interactive Calculator for Pythagorean Triples
• The Pythagorean Tree: A New Species [1] Compares the Barning Tree to an entirely new one.