Pythagorean triples are sets of three integers (a,b,c) where a2 + b2 = c2.
The simplest of these (where 'a', 'b' and 'c' are all greater than 0) is (3,4,5) as
9 + 16 = 25. There are clearly an unlimited number of these
triples as any multiple of (3,4,5), say (33,44,55), also meets the requirements:
(1089 + 1936 = 3025).
Ignoring these multiples, how many of these triples are there?
The image on the left is constructed by taking the first two terms (a
and b) an plotting them on squared paper. In this chart the 'c' value is
now the distance of the points from the origin (the centre).
Light blue points represent those values of 'a' and 'b' that have no common
factors. There are basically five of these - (3,4), (5,12), (7,24), (8,15)
and (20, 21) - with c values of (5,13,25,17 and 29). The 8 fold symmetry
and the other 35 light
blue points arise because if
a2 + b2 = c2
is true for (a,b) it is also true for (b,a), (−b,a), (−a,−b) etc.
The darker blue points represent the multiples of the basic points,
(6,8), (9,12), (10,24) etc.
Note: the images on this page omit the triple (0,0,0) and all the multiples and permutations
of (0,1,1).
I will call the triples where 'a' and 'b' have no common factors and
b > a > 0
the prime triples. So how many prime triples are there? The next image shows all the
triples (prime or otherwise) out to (320,320).
The prime ones are shown in light blue and their multiples darker. Each prime
triple generates a line of evenly spaced multiples, which are shown in darker
blue. The really obvious lines are generated by variations of (3,4). There are 57 prime
triples in this diagram, repeated 8 times through reflection and permutation. They are:
The simplest way to generate a sequence of triples is to notice that
difference between consecutive square numbers [1,4,9,16,...] generates the sequence of odd
numbers [3,5,7,9,...]. If the odd number is also a square [9,25,49,...] you have
a triple. And because two consecutive square numbers can't have any
common factors it must be prime. So we have already found a mechanism for generating an
unlimited number of prime triples.
The numbers in the first column of the table 1 (above) belong to
this sequence. We can create a generic formula for these triples as follows:
Although the above algorithm generates an unending sequence of prime triples it
doesn't generate them all. Of the 57 prime triples in table 1, only 12 follow this pattern.
A more general approach
Starting with c = b + n, where n is a positive integer;
So n × (2b + n) must be a square number.
To find solutons to this equation we look at the prime factors of n and (2b + n), starting with n.
Some prime factors will appear an odd number of times and others an even number. We collect all the
even ones into one list and all the odd ones into another. If a prime factor appears an odd number
of times more than 2, then one goes into the odd list and the remainder into the even list.
The product off all the even factors will be a square number and the product off all the odd ones
will have no repeated prime factors. This is a rather long winded way of demonstrating that all integers
can be factorised into the product of a square and a term with no repeating factors, so long as we allow
the posibility that the square number is 1.
For n × (2b + n) to be a square number, (2b + n) must be the product of 'k' and another
square number (y2). So
Where 'x', 'y' and 'k' are all integers. When 'n' is even, another factorisation is possible:
In the interactive element above, if you set y = (1+√2)x
(or as close as you can get), the 'a' and 'b' values will be similar.
Some examples
n = 1
When 'n' is 1, 'x' is 1 and 'k' is 1. So
For integer solutions y must be odd (1,3,5,...). This generates the triples [1,0,1], [3,4,5],
[5,12,13], etc.
n = 2
When 'n' is 2 we use the second factorisation based on, n = 2m. So 'x' is 1 and 'k' is 1, giving;
This generates the sequence
When 'y' is odd, y2±1 is even and all the terms can be divided by 2, so this formulation
only generates prime triples when 'y' is even, making 'a' is a multiple of 4. Eliminating the
permutation of (3,4,5), multiples of (0,1,1) and other multiples leaves the prime sequence:
n = 3
For n = 3 we use the original formulation where k = 3 and x = 1.
'a', 'b' and 'c' are all divisible by 3 so all triples generated are multiples of those generated
by n = 1. In general when k > 1 'a', 'b' and 'c' will have 'k' as a common factor so
the triple can't be prime.
The equations in the second column are equivalent to the ones in the first column, when
k is even. For instance setting k = 2 in the first column generates the same triple as
k = 1 in the second. So we can generate all triples using just the first set of
equations. All the prime triples can generated using k values of 1 or 2. When k is odd
x2 − y2
must be even.
Pythagoras.py contains code for
generating the triples and the diagrams that illustrate this page. It is a bit out of date
so the agorithms don't match exactly those decribed here, though they get the correct results.
A plot of the prime triples out to (3200x3200). Each pixel represents a 10x10
cell. This is quite interesting as most of the points appear to lie on curves
and there are 'special' points where three of these curves appear to intersect.
The same area with the non prime triples (red) included. The straight lines
are generated by multiples of the smaller triples like (3,4,5).
And finally the prime triples shown out to (32000,32000). Each pixel now
represents 100x100 points.
The (a, b) values from the following set of triples appear to lie on a curve. They can be divided into
two sub sets bases on whether 'a' is odd or even:
These two lists actually define two parabolas that are almost coincident.
If we take two points from one of the lists, say
(a1, b1)
and
(a2, b2),
we can construct the equation:
a neater formulation is
In the above example the points in the left column lie on the curve
b2 + 578a − 83521 = 0
and those on the right
b2 + 576a − 82944 = 0
These curves intersect the a-axis at 144.5 (172/2)
and 144 (122) respectively. The are nearly co-incident because 17/12
isn't a bad approximation for √2.
They are examples of the general curves:
b2 + 2m2a − m4 = 0 and
b2 + 4m2a − 4m4 = 0
where 'm' is a positive integer. These can be derived from the generating equations:
a = x × k × y and b = (y2 − x2) × (k/2).
Substituting x = a/ky into the second equation gives:
b = (y2 − a2 / k2y2) × (k/2).
which can be aranged to give
a2 + (2ky2)b −
k2y4 = 0.
The 2 equations above correspond to the cases where k = 1 and 2, and m = y (a and b have been swapped).
There are actually 4 sets of parabolas, the other three are generated through the transformations:
(a,b) ⇔ (a,−b),
(a,b) ⇔ (b,a) and
(a,b) ⇔ (b,−a),
representing reflections in the lines a = 0, a = b and a = −b.
The interactive Triples application can be used to visualise these sets
of parameters, for instance, this image:
Shows the set of parabolas for k = 1 with the prime triples superimposed. I've called these the
'West' parabolas as they extend infinitely to the left. This shows that about half the
prime triples fall on these lines. If we add in the k = 2 we see that now all the primes are accounted
for.
If we add in the "East" parabolas (generated by the transformation (a,b) ⇔ (a,−b))
we get the following image,
which shows that all the prime triples lie on the intersection of an East and a West parabola with the same
'k' value (1 or 2).
Interestingly, if we add in the other two sets of parabolas, North and South
(generated from (b,a) and (b,−a)) we now see that
the triples that were the intersection of 2 East/West parameters are also at the intersection of
two North/South parabolas of the opposite k value (1 ⇔ 2).
So we can locate all the prime triples using just one set of parabolas. You can verify this using the
interactive application. The intersections that don't have
prime triples do have non prime triples, which you can aslo verify using the application.
If we take two complex numbers (a1,b1)
and (a2,b2)
and multiply them together
we get a third complex number (a3,b3). If these numbers are interpreted as
vectors then the lengths (l) of the three vectors are related as
l3 = l1 x l2.
So if (a1,b1,l1) and (a2,b2,l2)
are Pythagorean triples so is (a3,b3,l3).
When multiplying complex numbers;
We can use this as another way of generating
unlimited series of triples, for instance the powers of (3,4) are;
(3 + 4i)n =
a
b
c
x
y
k
(3 + 4i)1
(3 + 4i)
3
4
5
1
3
1
(3 + 4i)2
(-7 + 24i)
7
24
25
1
7
1
(3 + 4i)3
(-117 + 44i)
44
117
125
2
11
2
(3 + 4i)4
(-527 - 336i)
336
527
625
7
24
2
(3 + 4i)5
(-237 - 3116i)
237
3,116
3,125
3
79
1
(3 + 4i)6
(11753 - 10296i)
10,296
11,753
15,625
44
117
2
(3 + 4i)7
(76443 + 16124i)
16,124
76,443
78,125
29
278
2
(3 + 4i)8
(164833 + 354144i)
164,833
354,144
390,625
191
863
1
(3 + 4i)9
(-922077 + 1721764i)
922,077
1,721,764
1,953,125
481
1917
1
(3 + 4i)10
(-9653287 + 1476984i)
1,476,984
9,653,287
9,765,625
237
3116
2
Although this series is unlimited the examples it generates are rather far
apart, only the first six appear in our diagrams above.
The 'x', 'y' and 'k' columns shows the values used to generate this triple, using
the first set of equations.
Notice that all the 'k' values are 1 or 2, so all these triples are prime, according to
our original definition. An alternative definition of primeness based on this complex
number mutiplication would render all but (3,4,5) composite.
The series generated by (5,12)
shoots out of sight even more rapidly.
a
b
c
x
y
k
5
12
13
1
5
1
119
120
132
7
17
1
828
2,035
133
3
138
2
239
28,560
134
1
239
1
145,668
341,525
135
122
1194
2
3,369,960
3,455,641
136
828
2035
2
23,161,315
58,317,492
137
2105
11003
1
13,651,680
815,616,479
138
239
28560
2
4,241,902,555
9,719,139,348
139
29755
142561
1
95,420,159,401
99,498,527,400
1310
195857
487193
1
When multiplying two different triples together there are two possible result
sets, for instance multiplying the 64 permutations of (3,4,5) and (5,12,13)
generates two sets of 32 values based on (16, 63, 65) and (33, 56, 65). For instance;
Generally, if 'c1' and 'c2'
are the largest values in different triples,
then c1 × c2
will be the largest value in two different
triples. This rule doesn't work when c1 = c2 as one of the triples
turns out to be (0,c,c).
If you have google Chrome with the native client plugin enabled you can use
this application
to calculate the triples in particlar ranges and generate images like those on this page.
This doesn't really work any more, I will replace it with a javascript version when I get a chance.
