A Pythagorean triple is a triple of positive integers ,
,
and
such that a right triangle exists with legs
and hypotenuse
. By the Pythagorean
theorem, this is equivalent to finding positive
integers
,
,
and
satisfying
|
(1)
|
The smallest and best-known Pythagorean triple is . The right triangle
having these side lengths is
sometimes called the 3, 4, 5 triangle.
The Old Babylonian tablet Plimpton 322, dated to roughly 1790-1750 BC, contains 15 rows associated with Pythagorean triples. Interpretations of the tablet's purpose include a teacher's aid for constructing right triangle problems and a table of numerical ratios; its intended use remains debated (Robson 2001).
Plots of points in the -plane such that
is a Pythagorean triple are shown above
for successively larger bounds. These plots include negative
values of
and
,
and are therefore symmetric about both the x-axis
and y-axis.
Similarly, plots of points in the -plane such that
is a Pythagorean triple are shown above
for successively larger bounds.
It is usual to consider only primitive Pythagorean triples (also called "reduced" triples) in which and
are relatively prime,
since other solutions can be generated trivially from the primitive ones. The primitive triples are illustrated above,
and it can be seen immediately that the radial lines
corresponding to imprimitive triples in the original plot are absent in this figure.
For primitive solutions, one of
or
must be even, and the other
odd (Shanks 1993, p. 141), with
always odd.
In addition, one side of every Pythagorean triple is divisible by 3, another by 4, and another by 5. One side may have two of these divisors, as in (8, 15, 17), (7, 24, 25), and (20, 21, 29), or even all three, as in (11, 60, 61).
Given a primitive triple with
odd and
even, three new primitive
triples are obtained from
|
(2)
| |||
|
(3)
| |||
|
(4)
|
where
|
(5)
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|
(6)
| |||
|
(7)
|
Hall (1970) and Roberts (1977) prove that is a primitive
Pythagorean triple with
odd and
even iff
|
(8)
|
where
is a possibly empty finite product
of the matrices
,
,
. It therefore follows that every primitive
Pythagorean triple with its odd leg
first must be a member of the infinite array
|
(9)
|
Pythagoras and the Babylonians gave a formula for generating (not necessarily primitive) triples as
|
(10)
|
for an integer , which generates a set of distinct triples containing
neither all primitive nor all imprimitive triples (and where in the special
case
,
).
The early Greeks gave
|
(11)
|
where
and
are positive integers with
, are relatively prime,
and have opposite parity (Shanks 1993, p. 141), which
generates a set of distinct triples containing precisely the primitive
triples (after appropriately sorting
and
).
Let
be a Fibonacci number and let
be a positive integer.
Then
|
(12)
|
generates distinct Pythagorean triples (Dujella 1995), although not exhaustively for either primitive or imprimitive triples. More generally, starting with positive
integers ,
,
and constructing the Fibonacci-like sequence
with terms
,
,
,
,
, ... generates distinct Pythagorean triples
|
(13)
|
(Horadam 1961), where
|
(14)
|
where
is a Lucas number.
For any Pythagorean triple, the product of the two nonhypotenuse legs (i.e., the two smaller numbers) is always divisible by 12, and the product of all three sides is divisible by 60. It is not known if there are two distinct triples having the same product. The existence of two such triples corresponds to a nonzero solution to the Diophantine equation
|
(15)
|
(Guy 1994, p. 188).
For a Pythagorean triple (,
,
),
|
(16)
|
where
counts the partitions of
into exactly three positive parts (Honsberger 1985). Every
three-term arithmetic progression of distinct
positive squares
,
,
, ordered with
, can be associated with a Pythagorean triple
by
|
(17)
| |||
|
(18)
| |||
|
(19)
|
(Robertson 1996).
The area of a triangle corresponding to the Pythagorean triple is
|
(20)
|
Fermat proved that a number of this form can never be a square number.
To find the number
of possible primitive triangles which may have
a leg (other than the hypotenuse)
of length
, write the positive integer
in the form
|
(21)
|
where the
are distinct primes and the
are positive integers.
The number of such triangles is then
|
(22)
|
i.e., 0 for
or singly even
and 2 to the power one less than the number of distinct
prime factors of
otherwise (Beiler 1966, pp. 115-116; Tripathi 2008/2009,
pp. 332-333). The first few numbers for
, 2, ..., are 0, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 2, 1, 0, 2,
... (OEIS A024361). To find the number of ways
in which a number
can be the leg (other than the hypotenuse)
of a primitive or nonprimitive right triangle,
write the factorization of
as
|
(23)
|
where the
are distinct odd primes,
is a nonnegative integer, and the
are positive integers.
Then
|
(24)
|
(Beiler 1966, p. 116; Tripathi 2008/2009, p. 334). Note that iff
is an odd prime
number or twice a prime number. The first few
numbers for
,
2, ... are 0, 0, 1, 1, 1, 1, 1, 2, 2, 1, 1, 4, 1, ... (OEIS A046079).
To find the number of ways in which a number
can be the hypotenuse of a
primitive right triangle, write its factorization
as
|
(25)
|
where the s
are distinct primes of
the form
and the
s
are distinct primes of
the form
,
is a nonnegative integer, and the other exponents are positive
integers. The number of possible primitive right
triangles is then
|
(26)
|
For example,
since
|
(27)
| |||
|
(28)
|
These primitive triples can also be constructed directly from the prime factorization
of the hypotenuse (Yekutieli 2023). If with distinct primes
and positive
integer exponents
, where
, choose Gaussian integers
with absolute
square
and form
|
(29)
|
Independently replace each by its complex conjugate,
keeping the first choice fixed to avoid conjugate duplication. The absolute
values of the real part and imaginary
part of
,
sorted into increasing order, give exactly the
primitive
triples with hypotenuse
. The unique factorization
of Gaussian integers ensures both primitiveness
and completeness. For
, the choices
and
give
or
, reproducing the two triples above.
The values of
for
,
2, ... are 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, ... (OEIS A024362).
The first few primes of
the form
are 5, 13, 17, 29, 37, 41, 53, 61, 73, 89, 97, 101, 109, 113, 137, ... (OEIS A002144),
so the smallest side lengths
which are the hypotenuses of 1, 2, 4, 8, 16, ... primitive right triangles are 5, 65, 1105,
32045, 1185665, 48612265, ... (OEIS A006278).
The number of possible primitive or nonprimitive right triangles having as a hypotenuse is
|
(30)
| |||
|
(31)
|
(correcting the typo of Beiler 1966, p. 117, which states that this formula gives the number of non-primitive solutions only), where is the sum of squares
function. For example, there are four distinct integer
triangles with hypotenuse 65, since
|
(32)
|
The first few numbers for , 2, ... are 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 1,
0, 1, 0, 0, ... (OEIS A046080). The smallest
hypotenuses having
distinct triples are 1, 5, 25, 125, 65, 3125, ... (OEIS A006339).
The following table gives the hypotenuses for which there exist exactly
distinct right integer triangles for
, 1, ..., 5.
| OEIS | hypotenuses
for which there exist | |
| 0 | A004144 | 1, 2, 3, 4, 6, 7, 8, 9, 11, 12, 14, 16, 18, ... |
| 1 | A084645 | 5, 10, 13, 15, 17, 20, 26, 29, 30, 34, 35, ... |
| 2 | A084646 | 25, 50, 75, 100, 150, 169, 175, 200, 225, ... |
| 3 | A084647 | 125, 250, 375, 500, 750, 875, 1000, 1125, 1375, ... |
| 4 | A084648 | 65, 85, 130, 145, 170, 185, 195, 205, 221, 255, ... |
| 5 | A084649 | 3125, 6250, 9375, 12500, 18750, 21875, 25000, ... |
Therefore, the total number of ways in which may be either a leg or hypotenuse
of a right triangle is given by
|
(33)
|
The values for ,
2, ... are 0, 0, 1, 1, 2, 1, 1, 2, 2, 2, 1, 4, 2, 1, 5, 3, ... (OEIS A046081).
The smallest numbers
which may be the sides of
general right triangles
for
,
2, ... are 3, 5, 16, 12, 15, 125, 24, 40, ... (OEIS A006593,
Beiler 1966, p. 114).
There are 50 Pythagorean triples with hypotenuse less than 100, the first few of which, sorted by increasing , are (3, 4, 5), (6, 8,10), (5, 12, 13), (9, 12, 15), (8, 15,
17), (12, 16, 20), (15, 20, 25), (7, 24, 25), (10, 24, 26), (20, 21, 29), (18, 24,
30), (16, 30, 34), (21, 28, 35), ... (OEIS A046083,
A046084, and A009000).
Of these, only 16 are primitive triplets with hypotenuse less than 100: (3, 4,5), (5, 12, 13), (8, 15, 17), (7, 24, 25), (20, 21, 29), (12, 35, 37), (9, 40, 41), (28, 45, 53), (11, 60, 61), (33, 56, 65), (16, 63, 65), (48, 55, 73), (36, 77, 85), (13, 84, 85), (39, 80, 89), and (65, 72, 97) (OEIS A046086, A046087, and A020882).
Let the number of triples with hypotenuse be denoted
, the number of triples with hypotenuse
be denoted
, and the number of primitive
triples less than
be denoted
. Then the following table summarizes the values for
powers of 10.
Lehmer (1900) proved that the number of primitive solutions with hypotenuse less than
satisfies
|
(34)
|
(OEIS A086201).
The inradii of the first few primitive Pythagorean triangles ordered by increasing are given by 1, 2, 3, 3, 6, 5, 4, 10, 5, ... (OEIS A014498).
There is a general method for obtaining triplets of Pythagorean triangles with equal areas. Take the three sets of generators,
for distinct positive integers and
, as
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(35)
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(36)
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(37)
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(38)
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(39)
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(40)
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Then the right triangle generated by each triple ()
has common area
|
(41)
|
(Beiler 1966, pp. 126-127). The smallest area shared by three distinct Pythagorean triangles is
attained at ,
which results in an area of 840 and corresponds to the triplets
(24, 70, 74), (40, 42, 58), and (15, 112, 113) (Beiler 1966, p. 126).
Right triangles whose areas consist of a single repeated digit include (area of 6) and
(area of 666666, Wells
1986, p. 89).
In 1643, Fermat challenged Mersenne to find a Pythagorean triplet whose hypotenuse and sum of the legs were squares. Fermat found the smallest such solution:
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(42)
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(43)
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(44)
|
with
|
(45)
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(46)
|
A related problem is to determine if a specified positive integer
can be the area of a right
triangle with rational side lengths. 1, 2, 3, and 4 are not the areas
of any rational-sided right
triangles, but 5 is (3/2, 20/3, 41/6), as is 6 (3, 4, 5). The solution to the
problem involves the elliptic curve
|
(47)
|
Such a right triangle exists iff (47) has a finite rational point with
. A triangle with side lengths (
,
,
) gives such a point by
|
(48)
| |||
|
(49)
|
(Koblitz 1993). There is no known general method for determining if there is a solution for arbitrary ,
but a technique devised by J. Tunnell in 1983 allows certain values to be ruled
out (Cipra 1996).