APythagorean triple consists of threepositive integersa,b, andc, such thata² +b² =c². Such a triple is commonly written(a,b,c), 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 integerk. A triangle whose side lengths are a Pythagorean triple is aright triangle and called aPythagorean triangle.

Aprimitive Pythagorean triple is one in whicha,b andc arecoprime (that is, they have no common divisor larger than 1).^[1] For example,(3, 4, 5) is a primitive Pythagorean triple whereas(6, 8, 10) is not. Every Pythagorean triple can be scaled to a unique primitive Pythagorean triple by dividing(a,b,c) by theirgreatest common divisor. Conversely, every Pythagorean triple can be obtained by multiplying the elements of a primitive Pythagorean triple by a positive integer (the same for the three elements).

The name is derived from thePythagorean theorem, stating that every right triangle has side lengths satisfying the formula${\displaystyle a^2+b^2=c^2}$; 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, thetriangle with sides${\displaystyle a=b=1}$ and${\displaystyle c=\sqrt{2}}$ is a right triangle, but${\displaystyle (1,1,\sqrt{2})}$ is not a Pythagorean triple because thesquare root of 2 is not an integer. Moreover,${\displaystyle 1}$ and${\displaystyle \sqrt{2}}$ do not have an integer common multiple because${\displaystyle \sqrt{2}}$ isirrational.

Pythagorean triples have been known since ancient times. The oldest known record comes fromPlimpton 322, a Babylonian clay tablet from about 1800 BC, written in asexagesimal number system.^[2]

When searching for integer solutions, theequationa² +b² =c² is aDiophantine equation. Thus Pythagorean triples are among the oldest known solutions of anonlinear Diophantine equation.

There are 16 primitive Pythagorean triples of numbers up to 100:

Other small Pythagorean triples such as (6, 8, 10) are not listed because they are not primitive; for instance (6, 8, 10) is a multiple of (3, 4, 5).

Each of these points (with their multiples) forms a radiating line in the scatter plot to the right.

Additionally, these are the remaining primitive Pythagorean triples of numbers up to 300:

Euclid's formula^[3] is a fundamental formula for generating Pythagorean triples given an arbitrary pair of integersm andn withm >n > 0. The formula states that the integers

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

form a Pythagorean triple. For example, given

${\displaystyle m=2,n=1}$ 

generate the primitive triple (3,4,5):

${\displaystyle a=2^2-1^2=3,b=2\times 2\times 1=4,c=2^2+1^2=5.}$ 

The triple generated byEuclid's formula is primitive if and only ifm andn arecoprime and exactly one of them is even. When bothm andn are odd, thena,b, andc will be even, and the triple will not be primitive; however, dividinga,b, andc by 2 will yield a primitive triple whenm andn are coprime.^[4]

Every primitive triple arises (after the exchange ofa andb, ifa is even) from aunique pair of coprime numbersm,n, one of which is even. It follows that there are infinitely many primitive Pythagorean triples. This relationship ofa,b andc tom andn 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—for example, (9, 12, 15) cannot be generated using integerm andn. This can be remedied by inserting an additional parameterk to the formula. The following will generate all Pythagorean triples uniquely:

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

wherem,n, andk are positive integers withm >n, and withm andn coprime and not both odd.

That these formulas generate Pythagorean triples can be verified by expandinga² +b² usingelementary algebra and verifying that the result equalsc². Since every Pythagorean triple can be divided through by some integerk to obtain a primitive triple, every triple can be generated uniquely by using the formula withm andn to generate its primitive counterpart and then multiplying through byk as in the last equation.

Choosingm andn from certain integer sequences gives interesting results. For example, ifm andn are consecutivePell numbers,a andb will differ by 1.^[5]

Many formulas for generating triples with particular properties have been developed since the time of Euclid.

That satisfaction of Euclid's formula bya, b, c issufficient for the triangle to be Pythagorean is apparent from the fact that for positive integersm andn,m >n, thea,b, andc given by the formula are all positive integers, and from the fact that

${\displaystyle a^2+b^2=(m^2-n^2{)}^2+(2mn{)}^2=(m^2+n^2{)}^2=c^2.}$ 

A proof of thenecessity thata, b, c be expressed by Euclid's formula for any primitive Pythagorean triple is as follows.^[6] All such primitive triples can be written as(a,b,c) wherea² +b² =c² anda,b,c arecoprime. Thusa,b,c arepairwise coprime (if a prime number divided two of them, it would be forced also to divide the third one). Asa andb are coprime, at least one of them is odd. If we suppose thata is odd, thenb is even andc is odd (if botha andb were odd,c would be even, andc² would be a multiple of 4, whilea² +b² would becongruent to 2 modulo 4, as an odd square is congruent to 1 modulo 4).

From${\displaystyle a^2+b^2=c^2,}$  assumea is odd. We obtain${\displaystyle c^2-a^2=b^2}$  and hence${\displaystyle (c-a)(c+a)=b^2.}$  Then${\displaystyle \frac{(c+a)}{b}=\frac{b}{(c-a)}.}$  Since${\displaystyle \frac{(c+a)}{b}}$  is rational, we set it equal to${\displaystyle \frac{m}{n}}$  in lowest terms. Thus${\displaystyle \frac{(c-a)}{b}=\frac{n}{m},}$  being the reciprocal of${\displaystyle \frac{(c+a)}{b}.}$  Then solving

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

for${\displaystyle \frac{c}{b}}$  and${\displaystyle \frac{a}{b}}$  gives

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

As${\displaystyle \frac{m}{n}}$  is fully reduced,m andn are coprime, and they cannot both be even. If they were both odd, the numerator of${\displaystyle \frac{m^2-n^2}{2mn}}$  would be a multiple of 4 (because an odd square is congruent to 1 modulo 4), and the denominator 2mn would not be a multiple of 4. Since 4 would be the minimum possible even factor in the numerator and 2 would be the maximum possible even factor in the denominator, this would implya to be even despite defining it as odd. Thus one ofm andn is odd and the other is even, and the numerators of the two fractions with denominator 2mn are odd. Thus these fractions are fully reduced (an odd prime dividing this denominator divides one ofm andn but not the other; thus it does not dividem² ±n²). One may thus equate numerators with numerators and denominators with denominators, giving Euclid's formula

${\displaystyle a=m^2-n^2,b=2mn,c=m^2+n^2}$  withm andn coprime and of opposite parities.

A longer but more commonplace proof is given in Maor (2007)^[7] and Sierpiński (2003).^[8] Another proof is given inDiophantine equation § Example of Pythagorean triples, as an instance of a general method that applies to everyhomogeneous Diophantine equation of degree two.

Suppose the sides of a Pythagorean triangle have lengthsm² −n²,2mn, andm² +n², and suppose the angle between the leg of lengthm² −n² and thehypotenuse of lengthm² +n² is denoted asβ. Then${\displaystyle \tan \frac{\beta }{2}=\frac{n}{m}}$  and the full-angle trigonometric values are${\displaystyle \sin \beta =\frac{2mn}{m^2+n^2}}$ ,${\displaystyle \cos \beta =\frac{m^2-n^2}{m^2+n^2}}$ , and⁠${\displaystyle \tan \beta =\frac{2mn}{m^2-n^2}}$ ⁠.^[9]

The following variant of Euclid's formula is sometimes more convenient, as being more symmetric inm andn (same parity condition onm andn).

Ifm andn are two odd integers such thatm >n, then

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

are three integers that form a Pythagorean triple, which is primitive if and only ifm andn are coprime. Conversely, every primitive Pythagorean triple arises (after the exchange ofa andb, ifa is even) from a unique pairm >n > 0 of coprime odd integers.

In the presentation above, it is said that all Pythagorean triples are uniquely obtained from Euclid's formula "after the exchange ofa andb, ifa is even". Euclid's formula and the variant above can be merged as follows to avoid this exchange, leading to the following result.

Every primitive Pythagorean triple can be uniquely written

${\displaystyle a=2\epsilon mn,b=\epsilon (m^2-n^2),c=\epsilon (m^2+n^2),}$ 

wherem andn are positive coprime integers, and${\displaystyle \epsilon =\frac{1}{2}}$  ifm andn are both odd, and${\displaystyle \epsilon =1}$  otherwise. Equivalently,${\displaystyle \epsilon =\frac{1}{2}}$  ifa is odd, and${\displaystyle \epsilon =1}$  ifa is even.

The properties of a primitive Pythagorean triple(a,b,c) witha <b <c (without specifying which ofa orb is even and which is odd) include:

• ${\displaystyle \frac{(c-a)(c-b)}{2}}$  is always a perfect square.^[10] As it is only a necessary condition but not a sufficient one, it can be used in checking if a given triple of numbers isnot a Pythagorean triple. For example, the triples{6, 12, 18} and{1, 8, 9} each pass the test that(c −a)(c −b)/2 is a perfect square, but neither is a Pythagorean triple.
• When a triple of numbersa,b andc forms a primitive Pythagorean triple, then(c minus the even leg) and one-half of(c minus the odd leg) are both perfect squares; however this is not a sufficient condition, as the numbers{1, 8, 9} pass the perfect squares test but are not a Pythagorean triple since1² + 8² ≠ 9².
• At most one ofa,b,c is a square.^[11]
• The area of a Pythagorean triangle cannot be the square^{[12]: p. 17} or twice the square^{[12]: p. 21} of an integer.
• Exactly one ofa,b isdivisible by 2 (iseven), and the hypotenusec is always odd.^[13]
• Exactly one ofa,b is divisible by 3, but neverc.^{[14][8]: 23–25}
• Exactly one ofa,b is divisible by 4,^[8] but neverc (becausec is never even).
• Exactly one ofa,b,c is divisible by 5.^[8]
• The largest number that always dividesabc is 60.^[15]
• Any odd number of the form2m+1, wherem is an integer andm>1, can be the odd leg of a primitive Pythagorean triple. Seealmost-isosceles primitive Pythagorean triples section below. However, only even numbers divisible by 4 can be the even leg of a primitive Pythagorean triple. This is becauseEuclid's formula for the even leg given above is2mn and one ofm orn must be even.
• The hypotenusec (which is always odd) is the sum of two squares. This requires all of its prime factors to beprimes of the form4n + 1.^[16] Therefore, c is of the form4n + 1. A sequence of possible hypotenuse numbers for a primitive Pythagorean triple can be found at (sequenceA008846 in theOEIS).
• The area(K =ab/2) is acongruent number^[17] divisible by 6.
• In every Pythagorean triangle, the radius of theincircle and the radii of the threeexcircles are positive integers. Specifically, for a primitive triple the radius of the incircle isr =n(m −n), and the radii of the excircles opposite the sidesm² −n²,2mn, and the hypotenusem² +n² are respectivelym(m −n),n(m +n), andm(m +n).^[18]
• As for any right triangle, the converse ofThales' theorem says that the diameter of thecircumcircle equals the hypotenuse; hence for primitive triples the circumdiameter ism² +n², and the circumradius is half of this and thus is rational but non-integer (sincem andn have opposite parity).
• When the area of a Pythagorean triangle is multiplied by thecurvatures of its incircle and 3 excircles, the result is four positive integersw >x >y >z, respectively. Integers−w,x,y,z satisfyDescartes's Circle Equation.^[19] Equivalently, the radius of theouter Soddy circle of any right triangle is equal to its semiperimeter. The outer Soddy center is located atD, whereACBD is a rectangle,ACB the right triangle andAB its hypotenuse.^{[19]: p. 6}
• Only two sides of a primitive Pythagorean triple can be simultaneously prime because byEuclid's formula for generating a primitive Pythagorean triple, one of the legs must be composite and even.^[20] However, only one side can be an integer of perfect power${\displaystyle p\ge 2}$  because if two sides were integers of perfect powers with equal exponent${\displaystyle p}$  it would contradict the fact that there are no integer solutions to theDiophantine equation${\displaystyle x^{2p}\pm y^{2p}=z^2}$ , with${\displaystyle x}$ ,${\displaystyle y}$  and${\displaystyle z}$  being pairwise coprime.^[21]
• There are no Pythagorean triangles in which the hypotenuse and one leg are the legs of another Pythagorean triangle; this is one of the equivalent forms ofFermat's right triangle theorem.^{[12]: p. 14}
• Each primitive Pythagorean triangle has a ratio of area,K, to squaredsemiperimeter,s, that is unique to itself and is given by^[22]

${\displaystyle \frac{K}{s^2}=\frac{n(m-n)}{m(m+n)}=1-\frac{c}{s}.}$ 

• No primitive Pythagorean triangle has an integer altitude from the hypotenuse; that is, every primitive Pythagorean triangle is indecomposable.^[23]
• The set of all primitive Pythagorean triples forms a rootedternary tree in a natural way; seeTree of primitive Pythagorean triples.
• Neither of theacute angles of a Pythagorean triangle can be arational number ofdegrees.^[24] (This follows fromNiven's theorem.)

In addition, special Pythagorean triples with certain additional properties can be guaranteed to exist:

• Every integer greater than 2 that is notcongruent to 2 mod 4 (in other words, every integer greater than 2 which isnot of the form4k + 2) is part of a primitive Pythagorean triple. (If the integer has the form4k, one may taken = 1 andm = 2k in Euclid's formula; if the integer is2k + 1, one may taken =k andm =k + 1.)
• 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 in which the hypotenuse and the longest leg differ by exactly one. Such triples are necessarily primitive and have the form(2n + 1, 2n² + 2n, 2n² + 2n +1). This results from Euclid's formula by remarking that the condition implies that the triple is primitive and must verify(m² +n²) - 2mn = 1. This implies(m –n)² = 1, and thusm =n + 1. The above form of the triples results thus of substitutingm forn + 1 in Euclid's formula.
• There exist infinitely many primitive Pythagorean triples in which the hypotenuse and the longest leg differ by exactly two. They are all primitive, and are obtained by puttingn = 1 in Euclid's formula. More generally, for every integerk > 0, there exist infinitely many primitive Pythagorean triples in which the hypotenuse and the odd leg differ by2k². They are obtained by puttingn =k in Euclid's formula.
• There exist infinitely many Pythagorean triples in which the two legs differ by exactly one. For example, 20² + 21² = 29²; these are generated by Euclid's formula when${\displaystyle \frac{m-n}{n}}$  is aconvergent to${\displaystyle \sqrt{2}.}$ 
• For each positive integerk, there existk Pythagorean triples with different hypotenuses and the same area.
• For each positive integerk, there exist at leastk different primitive Pythagorean triples with the same lega, wherea is some positive integer (the length of the even leg is 2mn, and it suffices to choosea with many factorizations, for examplea = 4b, whereb is a product ofk different odd primes; this produces at least2^k different primitive triples).^[8]: 30
• For each positive integerk, there exist at leastk different Pythagorean triples with the same hypotenuse.^[8]: 31
• Ifc =p^e is aprime power, there exists a primitive Pythagorean triplea² +b² =c² if and only if the primep has the form4n + 1; this triple is uniqueup to the exchange ofa andb.
• More generally, a positive integerc is the hypotenuse of a primitive Pythagorean triple if and only if eachprime factor ofc iscongruent to1 modulo4; that is, each prime factor has the form4n + 1. In this case, the number of primitive Pythagorean triples(a,b,c) witha <b is2^k−1, wherek is the number of distinct prime factors ofc.^[25]
• There exist infinitely many Pythagorean triples with square numbers for both the hypotenusec and the sum of the legsa +b. According to Fermat, thesmallest such triple^[26] has sidesa = 4,565,486,027,761;b = 1,061,652,293,520; andc = 4,687,298,610,289. Herea +b = 2,372,159² andc = 2,165,017². This is generated by Euclid's formula with parameter valuesm = 2,150,905 andn = 246,792.
• There exist non-primitivePythagorean triangles with integer altitude from the hypotenuse.^[27][28] Such Pythagorean triangles are known asdecomposable since they can be split along this altitude into two separate and smaller Pythagorean triangles.^[23]

Euclid's formula for a Pythagorean triple

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

can be understood in terms of the geometry ofrational points on theunit circle (Trautman 1998).

In fact, a point in theCartesian plane with coordinates(x,y) belongs to the unit circle ifx² +y² = 1. The point isrational ifx andy arerational numbers, that is, if there arecoprime integersa,b,c such that

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

By multiplying both members byc², one can see that the rational points on the circle are in one-to-one correspondence with the primitive Pythagorean triples.

The unit circle may also be defined by aparametric equation

${\displaystyle x=\frac{1-t^2}{1+t^2},y=\frac{2t}{1+t^2}.}$ 

Euclid's formula for Pythagorean triples and the inverse relationshipt =y / (x + 1) mean that, except for(−1, 0), a point(x,y) on the circle is rational if and only if the corresponding value oft is a rational number. Note thatt =y / (x + 1) =b / (a +c) =n /m is also thetangent of half of the angle that is opposite the triangle side of lengthb.

There is a correspondence betweenpoints on the unit circle with rational coordinates and primitive Pythagorean triples. At this point, Euclid's formulae can be derived either by methods oftrigonometry or equivalently by using thestereographic projection.

For the stereographic approach, suppose thatP′ is a point on thex-axis with rational coordinates

${\displaystyle P^{\prime }=\left(\frac{m}{n},0\right).}$ 

Then, it can be shown by basic algebra that the pointP has coordinates

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

This establishes that eachrational point of thex-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 thex-axis, follows by applying the inverse stereographic projection. Suppose thatP(x,y) is a point of the unit circle withx andy rational numbers. Then the pointP′ obtained by stereographic projection onto thex-axis has coordinates

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

which is rational.

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

A 2Dlattice is a regular array of isolated points where if any one point is chosen as the Cartesian origin (0, 0), then all the other points are at(x,y) wherex andy range over all positive and negative integers. Any Pythagorean triangle with triple(a,b,c) can be drawn within a 2D lattice with vertices at coordinates(0, 0),(a, 0) and(0,b). The count of lattice points lying strictly within the bounds of the triangle is given by  ${\displaystyle \frac{(a-1)(b-1)-gcd(a,b)+1}{2};}$ ^[29] for primitive Pythagorean triples this interior lattice count is  ${\displaystyle \frac{(a-1)(b-1)}{2}.}$  The area (byPick's theorem equal to one less than the interior lattice count plus half the boundary lattice count) equals  ${\displaystyle \frac{ab}{2}}$  .

The first occurrence of two primitive Pythagorean triples sharing the same area occurs with triangles with sides(20, 21, 29), (12, 35, 37) and common area 210 (sequenceA093536 in theOEIS). The first occurrence of two primitive Pythagorean triples sharing the same interior lattice count occurs with(18108, 252685, 253333), (28077, 162964, 165365) and interior lattice count 2287674594 (sequenceA225760 in theOEIS). Three primitive Pythagorean triples have been found sharing the same area:(4485, 5852, 7373),(3059, 8580, 9109),(1380, 19019, 19069) with area 13123110. As yet, no set of three primitive Pythagorean triples have been found sharing the same interior lattice count.

By Euclid's formula all primitive Pythagorean triples can be generated from integers${\displaystyle m}$  and${\displaystyle n}$  with${\displaystyle m>n>0}$ ,${\displaystyle m+n}$  odd and${\displaystyle gcd(m,n)=1.}$  Hence there is a 1 to 1 mapping of rationals (in lowest terms) to primitive Pythagorean triples where${\displaystyle \frac{n}{m}}$  is in the interval${\displaystyle (0,1)}$  and${\displaystyle m+n}$  odd.

The reverse mapping from a primitive triple${\displaystyle (a,b,c)}$  where${\displaystyle c>b>a>0}$  to a rational${\displaystyle \frac{n}{m}}$  is achieved by studying the two sums${\displaystyle a+c}$  and${\displaystyle b+c.}$  One of these sums will be a square that can be equated to${\displaystyle (m+n{)}^2}$  and the other will be twice a square that can be equated to${\displaystyle 2m^2.}$  It is then possible to determine the rational${\displaystyle \frac{n}{m}.}$ 

In order to enumerate primitive Pythagorean triples the rational can be expressed as an ordered pair${\displaystyle (n,m)}$  and mapped to an integer using a pairing function such asCantor's pairing function. An example can be seen at (sequenceA277557 in theOEIS). It begins

${\displaystyle 8,18,19,32,33,34,\dots }$  and gives rationals

${\displaystyle \frac{1}{2},\frac{2}{3},\frac{1}{4},\frac{3}{4},\frac{2}{5},\frac{1}{6},\dots }$  these, in turn, generate primitive triples

${\displaystyle (3,4,5),(5,12,13),(8,15,17),(7,24,25),(20,21,29),(12,35,37),\dots }$ 

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

 

A matrix of this form issymmetric. Furthermore, thedeterminant ofX is

${\displaystyle detX=c^2-a^2-b^2}$ 

which is zero precisely when(a,b,c) is a Pythagorean triple. IfX corresponds to a Pythagorean triple, then as a matrix it must haverank 1.

SinceX is symmetric, it follows from a result inlinear algebra that there is acolumn vectorξ = [mn]^T such that theouter product

holds, where theT denotes thematrix transpose. Since ξ and -ξ produce the same Pythagorean triple, the vector ξ can be considered aspinor (for theLorentz group SO(1, 2)). In abstract terms, the Euclid 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).

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

 

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

 

whereu andv are selected (by theEuclidean algorithm) so thatmu +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 ifA is a matrix inΓ, then

gives rise to an action on the matrixX 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 (perTrautman 1998) to call a triple(a,b,c)standard ifc > 0 and either(a,b,c) are relatively prime or(a/2,b/2,c/2) are relatively prime witha/2 odd. If the spinor[mn]^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 ofm andn for whichm is odd andn is even. Let thesubgroup Γ(2) of Γ be thekernel of thegroup homomorphism

${\displaystyle \mathrm{\Gamma }=\mathrm{S}\mathrm{L}(2,\mathbf{Z})\to \mathrm{S}\mathrm{L}(2,{\mathbf{Z}}_2)}$ 

whereSL(2,Z₂) is thespecial linear group over thefinite fieldZ₂ ofintegers 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 ofAξ for allA ∈ 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 thefree group whose generators are the matrices

 

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

By a result ofBerggren (1934), all primitive Pythagorean triples can be generated from the (3, 4, 5) triangle by using the threelinear transformations T₁, T₂, T₃ below, wherea,b,c are sides of a triple:

In other words, every primitive triple will be a "parent" to three additional primitive triples.Starting from the initial node witha = 3,b = 4, andc = 5, the operationT₁ produces the new triple

(3 − (2×4) + (2×5), (2×3) − 4 + (2×5), (2×3) − (2×4) + (3×5)) = (5, 12, 13),

and similarlyT₂ andT₃ produce the triples (21, 20, 29) and (15, 8, 17).

The linear transformations T₁, T₂, and T₃ have a geometric interpretation in the language ofquadratic forms. They are closely related to (but are not equal to) reflections generating theorthogonal group ofx² +y² −z² over the integers.^[30]

Alternatively, Euclid's formulae can be analyzed and proved using theGaussian integers.^[31] Gaussian integers arecomplex numbers of the formα =u +vi, whereu andv are ordinaryintegers andi is thesquare root of negative one. Theunits of Gaussian integers are ±1 and ±i. The ordinary integers are called therational integers and denoted as 'Z'. The Gaussian integers are denoted asZ[i]. The right-hand side of thePythagorean theorem may be factored in Gaussian integers:

${\displaystyle c^2=a^2+b^2=(a+bi)\overline{(a+bi)}=(a+bi)(a-bi).}$ 

A primitive Pythagorean triple is one in whicha andb arecoprime, i.e., they share no prime factors in the integers. For such a triple, eithera orb is even, and the other is odd; from this, it follows thatc is also odd.

The two factorsz :=a +bi andz* :=a −bi of a primitive Pythagorean triple each equal the square of a Gaussian integer. This can be proved using the property that every Gaussian integer can be factored uniquely into Gaussian primesup tounits.^[32] (This unique factorization follows from the fact that, roughly speaking, a version of theEuclidean algorithm can be defined on them.) The proof has three steps. First, ifa andb share no prime factors in the integers, then they also share no prime factors in the Gaussian integers. (Assumea =gu andb =gv with Gaussian integersg,u andv andg not a unit. Thenu andv lie on the same line through the origin. All Gaussian integers on such a line are integer multiples of some Gaussian integerh. But then the integergh ≠ ±1 divides botha andb.) Second, it follows thatz andz* likewise share no prime factors in the Gaussian integers. For if they did, then their common divisorδ would also dividez +z* = 2a andz −z* = 2ib. Sincea andb are coprime, that implies thatδ divides 2 = (1 + i)(1 − i) = i(1 − i)². From the formulac² =zz*, that in turn would imply thatc is even, contrary to the hypothesis of a primitive Pythagorean triple. Third, sincec² is a square, every Gaussian prime in its factorization is doubled, i.e., appears an even number of times. Sincez andz* share no prime factors, this doubling is also true for them. Hence,z andz* are squares.

Thus, the first factor can be written

${\displaystyle a+bi=\epsilon {\left(m+ni\right)}^2,\epsilon \in \{\pm 1,\pm i\}.}$ 

The real and imaginary parts of this equation give the two formulas:

 

For any primitive Pythagorean triple, there must be integersm andn such that these two equations are satisfied. Hence, every Pythagorean triple can be generated from some choice of these integers.

If we consider the square of a Gaussian integer we get the following direct interpretation of Euclid's formula as representing the perfect square of a Gaussian integer.

${\displaystyle (m+ni{)}^2=(m^2-n^2)+2mni.}$ 

Using the facts that the Gaussian integers are a Euclidean domain and that for a Gaussian integer p${\displaystyle |p{|}^2}$  is always a square it is possible to show that a Pythagorean triple corresponds to the square of a prime Gaussian integer if the hypotenuse is prime.

If the Gaussian integer is not prime then it is the product of two Gaussian integers p and q with${\displaystyle |p{|}^2}$  and${\displaystyle |q{|}^2}$  integers. Since magnitudes multiply in the Gaussian integers, the product must be${\displaystyle |p||q|}$ , which when squared to find a Pythagorean triple must be composite. The contrapositive completes the proof.

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 appear in the plot, all integer multiples of(a,b) must also appear in the plot, and this property produces the appearance of lines radiating from the origin in the diagram.

Within the scatter, there are sets ofparabolic patterns with a high density of points and all their foci at the origin, opening up in all four directions. Different parabolas intersect at the axes and appear to reflect off the axis with an incidence angle of 45 degrees, with a third parabola entering in a perpendicular fashion. Within this quadrant, each arc centered on the origin shows that section of the parabola that lies between its tip and its intersection with itssemi-latus rectum.

These patterns can be explained as follows. If${\displaystyle a^2/4n}$  is an integer, then (a,${\displaystyle |n-a^2/4n|}$ ,${\displaystyle n+a^2/4n}$ ) is a Pythagorean triple. (In fact every Pythagorean triple(a,b,c) can be written in this way with integern, possibly after exchanginga andb, since${\displaystyle n=(b+c)/2}$  anda andb cannot both be odd.) The Pythagorean triples thus lie on curves given by${\displaystyle b=|n-a^2/4n|}$ , that is, parabolas reflected at thea-axis, and the corresponding curves witha andb interchanged. Ifa is varied for a givenn (i.e. on a given parabola), integer values ofb occur relatively frequently ifn is a square or a small multiple of a square. If several such values happen to lie close together, the corresponding parabolas approximately coincide, and the triples cluster in a narrow parabolic strip. For instance,38² = 1444,2 × 27² = 1458,3 × 22² = 1452,5 × 17² = 1445 and10 × 12² = 1440; the corresponding parabolic strip aroundn ≈ 1450 is clearly visible in the scatter plot.

The angular properties described above follow immediately from the functional form of the parabolas. The parabolas are reflected at thea-axis ata = 2n, and the derivative ofb with respect toa at this point is –1; hence the incidence angle is 45°. Since the clusters, like all triples, are repeated at integer multiples, the value2n also corresponds to a cluster. The corresponding parabola intersects theb-axis at right angles atb = 2n, and hence its reflection upon interchange ofa andb intersects thea-axis at right angles ata = 2n, precisely where the parabola forn is reflected at thea-axis. (The same is of course true fora andb interchanged.)

Albert Fässler and others provide insights into the significance of these parabolas in the context of conformal mappings.^[33][34]

The casen = 1 of the more general construction of Pythagorean triples has been known for a long time.Proclus, in his commentary to the47th Proposition of the first book ofEuclid'sElements, 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 toPythagoras. (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):

${\displaystyle \text{side }a:\text{side }b=\frac{a^2-1}{2}:\text{side }c=\frac{a^2+1}{2}.}$ 

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

${\displaystyle \text{side }a:\text{side }b={\left(\frac{a}{2}\right)}^2-1:\text{side }c={\left(\frac{a}{2}\right)}^2+1}$ 

It can be shown that all Pythagorean triples can be obtained, with appropriate rescaling, from the basic Platonic sequence (a,(a² − 1)/2 and(a² + 1)/2) by allowinga to take non-integer rational values. Ifa is replaced with the fractionm/n in the sequence, the result is equal to the 'standard' triple generator (2mn,m² −n²,m² +n²) after rescaling. It follows that every triple has a corresponding rationala value which can be used to generate asimilar triangle (one with the same three angles and with sides in the same proportions as the original). For example, the Platonic equivalent of(56, 33, 65) is generated bya =m/n = 7/4 as(a, (a² –1)/2, (a²+1)/2) = (56/32, 33/32, 65/32). The Platonic sequence itself can be derived^{[clarification needed]} by following the steps for 'splitting the square' described inDiophantus II.VIII.

The equation,

${\displaystyle a^4+b^4+c^4+d^4=(a+b+c+d{)}^4}$ 

is equivalent to the special Pythagorean triple,

${\displaystyle (a^2+ab+b^2{)}^2+(c^2+cd+d^2{)}^2=((a+b{)}^2+(a+b)(c+d)+(c+d{)}^2{)}^2}$ 

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

${\displaystyle a,b,c,d=-2634,955,1770,5400}$ 

${\displaystyle a,b,c,d=-31764,7590,27385,48150}$ 

One way to generate solutions to${\displaystyle a^2+b^2=c^2+d^2}$  is to parametrizea, b, c, d in terms of integersm, n, p, q as follows:^[35]

${\displaystyle (m^2+n^2)(p^2+q^2)=(mp-nq{)}^2+(np+mq{)}^2=(mp+nq{)}^2+(np-mq{)}^2.}$ 

Given two sets of Pythagorean triples,

${\displaystyle (a^2-b^2{)}^2+(2ab{)}^2=(a^2+b^2{)}^2}$ 

${\displaystyle (c^2-d^2{)}^2+(2cd{)}^2=(c^2+d^2{)}^2}$ 

the problem of finding equal products of anon-hypotenuse side and the hypotenuse,

${\displaystyle (a^2-b^2)(a^2+b^2)=(c^2-d^2)(c^2+d^2)}$ 

is easily seen to be equivalent to the equation,

${\displaystyle a^4-b^4=c^4-d^4}$ 

and was first solved by Euler as${\displaystyle a,b,c,d=133,59,158,134.}$  Since he showed this is a rational point in anelliptic curve, then there is an infinite number of solutions. In fact, he also found a 7th degree polynomial parameterization.

For the case ofDescartes' circle theorem where all variables are squares,

${\displaystyle 2(a^4+b^4+c^4+d^4)=(a^2+b^2+c^2+d^2{)}^2}$ 

Euler showed this is equivalent to three simultaneous Pythagorean triples,

${\displaystyle (2ab{)}^2+(2cd{)}^2=(a^2+b^2-c^2-d^2{)}^2}$ 

${\displaystyle (2ac{)}^2+(2bd{)}^2=(a^2-b^2+c^2-d^2{)}^2}$ 

${\displaystyle (2ad{)}^2+(2bc{)}^2=(a^2-b^2-c^2+d^2{)}^2}$ 

There is also an infinite number of solutions, and for the special case when${\displaystyle a+b=c}$ , then the equation simplifies to,

${\displaystyle 4(a^2+ab+b^2)=d^2}$ 

with small solutions as${\displaystyle a,b,c,d=3,5,8,14}$  and can be solved asbinary quadratic forms.

No Pythagorean triples areisosceles, because the ratio of the hypotenuse to either other side is√2, but√2 cannot be expressed as the ratio of 2 integers.

There are, however,right-angled triangles with integral sides for which the lengths of thenon-hypotenuse sides differ by one, such as,

${\displaystyle 3^2+4^2=5^2}$ 

${\displaystyle {20}^2+{21}^2={29}^2}$ 

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

${\displaystyle {\left(\frac{x-1}{2}\right)}^2+{\left(\frac{x+1}{2}\right)}^2=y^2}$ 

where {x, y} are the solutions to thePell equation${\displaystyle x^2-2y^2=-1.}$ 

Ifa,b,c are the sides of this type of primitive Pythagorean triple then the solution to the Pell equation is given by therecursive formula

${\displaystyle a_n=6a_{n-1}-a_{n-2}+2}$  with${\displaystyle a_1=3}$  and${\displaystyle a_2=20}$ 

${\displaystyle b_n=6b_{n-1}-b_{n-2}-2}$  with${\displaystyle b_1=4}$  and${\displaystyle b_2=21}$ 

${\displaystyle c_n=6c_{n-1}-c_{n-2}}$  with${\displaystyle c_1=5}$  and⁠${\displaystyle c_2=29}$ ⁠.^[36]

This sequence of primitive Pythagorean triples forms the central stem (trunk) of therooted ternary tree of primitive Pythagorean triples.

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

${\displaystyle 5^2+{12}^2={13}^2}$ 

${\displaystyle 7^2+{24}^2={25}^2}$ 

then the complete solution for the primitive Pythagorean triplea,b,c is

${\displaystyle a=2m+1,b=2m^2+2m,c=2m^2+2m+1}$ 

and

${\displaystyle (2m+1{)}^2+(2m^2+2m{)}^2=(2m^2+2m+1{)}^2}$ 

where integer${\displaystyle m>0}$  is the generating parameter.

It shows that allodd 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

${\displaystyle a^b+b^a=Kc}$ 

for some integer${\displaystyle K}$ . Or in other words${\displaystyle a^b+b^a}$  is divisible by${\displaystyle c}$  such as in

${\displaystyle (5^{12}+{12}^5)/13=18799189}$ .^[37]

Starting with 5, every secondFibonacci 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$${\displaystyle (F_n{F}_{n+3}{)}^2+(2F_{n+1}{F}_{n+2}{)}^2=F_{2n+3}^2.}$$ 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.^[38]

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

The expression

${\displaystyle {\left(m_1^2-m_2^2-\dots -m_n^2\right)}^2+\sum _{k=2}^n(2m_1{m}_k{)}^2={\left(m_1^2+\dots +m_n^2\right)}^2}$ 

is a Pythagoreann-tuple for any tuple of positive integers(m₁, ...,m_n) withm²
₁ >m²
₂ + ... +m²
_n. The Pythagoreann-tuple can be made primitive by dividing out by the largest common divisor of its values.

Furthermore, any primitive Pythagoreann-tuplea²
₁ + ... +a²
_n =c² can be found by this approach. Use(m₁, ...,m_n) = (c +a₁,a₂, ...,a_n) to get a Pythagoreann-tuple by the above formula and divide out by the largest common integer divisor, which is2m₁ = 2(c +a₁). Dividing out by the largest common divisor of these(m₁, ...,m_n) values gives the same primitive Pythagoreann-tuple; and there is a one-to-one correspondence between tuples ofsetwise coprime positive integers(m₁, ...,m_n) satisfyingm²
₁ >m²
₂ + ... +m²
_n and primitive Pythagoreann-tuples.

Examples of the relationship between setwise coprime values${\displaystyle \overrightarrow{m}}$  and primitive Pythagoreann-tuples include:^[39]

 

Since the sumF(k,m) ofk consecutive squares beginning withm² is given by the formula,^[40]

${\displaystyle F(k,m)=km(k-1+m)+\frac{k(k-1)(2k-1)}{6}}$ 

one may find values(k,m) so thatF(k,m) is a square, such as one by Hirschhorn where the number of terms is itself a square,^[41]

${\displaystyle m=\frac{v^4-24v^2-25}{48},k=v^2,F(m,k)=\frac{v^5+47v}{48}}$ 

andv ≥ 5 is any integer not divisible by 2 or 3. For the smallest casev = 5, hencek = 25, this yields the well-known cannonball-stacking problem ofLucas,

${\displaystyle 0^2+1^2+2^2+\cdots +{24}^2={70}^2}$ 

a fact which is connected to theLeech lattice.

In addition, if in a Pythagoreann-tuple (n ≥ 4) alladdends are consecutive except one, one can use the equation,^[42]

${\displaystyle F(k,m)+p^2=(p+1{)}^2}$ 

Since the second power ofp cancels out, this is only linear and easily solved for as${\displaystyle p=\frac{F(k,m)-1}{2}}$  thoughk,m should be chosen so thatp is an integer, with a small example beingk = 5,m = 1 yielding,

${\displaystyle 1^2+2^2+3^2+4^2+5^2+{27}^2={28}^2}$ 

Thus, one way of generating Pythagoreann-tuples is by using, for variousx,^[43]

${\displaystyle x^2+(x+1{)}^2+\cdots +(x+q{)}^2+p^2=(p+1{)}^2,}$ 

whereq = n–2 and where

${\displaystyle p=\frac{(q+1)x^2+q(q+1)x+\frac{q(q+1)(2q+1)}{6}-1}{2}.}$ 

A generalization of the concept of Pythagorean triples is the search for triples of positive integersa,b, andc, such thataⁿ +bⁿ =cⁿ, for somen strictly greater than 2.Pierre de Fermat in 1637 claimed that no such triple exists, a claim that came to be known asFermat's Last Theorem because it took longer than any other conjecture by Fermat to be proved or disproved. The first proof was given byAndrew Wiles in 1994.

Another generalization is searching for sequences ofn + 1 positive integers for which thenth power of the last is the sum of thenth powers of the previous terms. The smallest sequences for known values ofn 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}

For then = 3 case, in which${\displaystyle x^3+y^3+z^3=w^3,}$  called theFermat cubic, a general formula exists giving all solutions.

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 withRamanujan about the number1729 being the smallest number that can be expressed as a sum of two cubes in two distinct ways.

There can also existn − 1 positive integers whosenth powers sum to annth power (though, byFermat's Last Theorem, not forn = 3); these are counterexamples toEuler's sum of powers conjecture. The smallest known counterexamples are^[44][45][15]

• n = 4: (95800, 217519, 414560; 422481)
• n = 5: (27, 84, 110, 133; 144)

AHeronian triangle is commonly defined as one with integer sides whose area is also an integer. The lengths of the sides of such a triangle form aHeronian triple(a, b, c) fora ≤b ≤c.Every Pythagorean triple is a Heronian triple, because at least one of the legsa,b must be even in a Pythagorean triple, so the areaab/2 is an integer. Not every Heronian triple is a Pythagorean triple, however, as the example(4, 13, 15) with area 24 shows.

If(a,b,c) is a Heronian triple, so is(ka,kb,kc) wherek is any positive integer; its area will be the integer that isk² times the integer area of the(a,b,c) triangle.The Heronian triple(a,b,c) isprimitive provideda,b,c aresetwise coprime. (With primitive Pythagorean triples the stronger statement that they arepairwise coprime also applies, but with primitive Heronian triangles the stronger statement does not always hold true, such as with(7, 15, 20).) Here are a few of the simplest primitive Heronian triples that are not Pythagorean triples:

(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

ByHeron's formula, the extra condition for a triple of positive integers(a,b,c) witha <b <c to be Heronian 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.

Primitive Pythagorean triples have been used in cryptography as random sequences and for the generation of keys.^[46]