
Pythagorean Triples, a^{2} + b^{2} = c^{2} Bill Richardson
This note is an examination of some different ways of generating Pythagorean triples.
The standard method used for obtaining primitive Pythagorean triples is to use the generating equations,
a = r^{2}  s^{2}, b = 2rs , c = r^{2} + s ^{2} (1)
where 0 < s < r , (r, s) = 1 with r and s of opposite parity.
Below are some primitive Pythagorean triples generated by the above equations.
 s = 1  

 a  b  c 
r = 2  3  4  5 
r = 4  15  8  17 
r = 6  35  12  37 
r = 8  63  16  65 
r = 10  99  20  101 
r = 12  143  24  145 
r = 14  195  28  197 
r = 16  255  32  257 
 s = 2  

 a  b  c 
r = 3  5  12  13 
r = 5  21  20  29 
r = 7  45  28  53 
r = 9  77  36  85 
r = 11  117  44  125 
r = 13  165  52  173 
r = 15  221  60  229 
r = 17  285  68  293 
From these few calculations we can observe some interesting patterns. For example, all the triples for which s = 1 we have that a and c are consecutive odd integers. Furthermore, the triples [3 4 5] and [5 12 13] have b and c as consecutive integers. And, finally, [3 4 5] and [21 20 29] have a and b as consecutive integers. Can we determine conditions that will allow us to generate only those types of triples we wish?
Conditions for a and c to be consecutive odd integers:
In formulas (1), if s = 1, then r must be an even integer of the form r = 2n .
We then have
a = 4n^{2}  1, b = 4n, c = 4n^{2} + 1 , n = 1, 2, 3, 4,... (2)
Or, we could write a and c as functions of b if we let b be an integer of the form b = 4n . Then
b = 4n , a = bn  1 , and c = bn + 1, n = 1, 2, 3, 4, ...
Conditions for b and c to be consecutive integers:
If b, c are to be consecutive integers, we have from (1) that r^{2} + s^{2} = 2rs + 1 . Thus, r^{2} + s^{2}  2rs + 1 and (r + s)^{2} =1 . Since 0 < s < r, r  s = 1 , or r = s + 1 . When we put these values of r in (1), we have
a = 2s + 1, b = 2s(s + 1), and c = 2s(s + 1) + 1, s = 1,2,3,4,... (3)
This result can also be written as
a = 2n + 1, b = (a + 1)n, and c = (a + 1)n + 1, n = 1,2,3,4,... (4)
Conditions for a and b to be consecutive integers:
This is a tougher problem. We do not get nice things happening, as in the previous cases. However, if we try to see if we can express the relationship between these triples as a recurrence relation, we see that the best chance is if there is a matrix
 a_{11}  a_{12}  a_{13}  
a_{21}  a_{22}  a_{23} 
a_{31}  a_{32}  a_{33} 
such that

3  4  5  
20  21  29 
119  120  169 
 * 
 a_{11}  a_{12}  a_{13}  
a_{21}  a_{22}  a_{23} 
a_{31}  a_{32}  a_{33} 
 = 
 20  21  29  . 
119  120  169 
696  697  985 

The solution to this matrix equation yields the matrix
 2  1  2  . 
1  2  2 
2  2  3 
Thus,
[ 3 4 5] 
* 
 2  1  2  
1  2  2 
2  2  3 
 =   [ 20 21 29] 

and
[20 21 29] 
* 
 2  1  2  
1  2  2 
2  2  3 
 =   [119 120 169] 

We therefore have a generator for the triples with a and b consecutive integers.
[ a_{n} b_{n} c_{n} ]
 =  [ a_{n1} b_{n1} c_{n1} ] 
* 
 2  1  2  
1  2  2 
2  2  3 


where a_{0} = 3, b_{0} = 4, c_{0} = 5.  (5) 
A natural question to ask now is, "Are there matrix forms for the previous two cases?" As it turns out there are. For a and c consecutive odd integers we need to solve the following

3  4  5  
15  8  17 
35  12  37 
 * 
 a_{11}  a_{12}  a_{13}  
a_{21}  a_{22}  a_{23} 
a_{31}  a_{32}  a_{33} 
 = 
 15  8  17  . 
35  12  37 
63  16  65 

The solution to this matrix equation yields the matrix
 1  2  2  . 
2  1  2 
2  2  3 
so that our matrix solution would be
[ a_{n} b_{n} c_{n} ]
 =  [ a_{n1} b_{n1} c_{n1} ] 
* 
 1  2  2  
2  1  2 
2  2  3 


where a_{0} = 3, b_{0} = 4, c_{0} = 5.  (6) 
For the case where b and c are consecutive integers, we need to solve the equation

3  4  5  
5  12  13 
7  24  25 
 * 
 a_{11}  a_{12}  a_{13}  
a_{21}  a_{22}  a_{23} 
a_{31}  a_{32}  a_{33} 
 = 
 5  12  13  . 
7  24  25 
9  40  41 

The solution to this matrix equation yields the matrix
 1  2  2  . 
2  1  2 
2  2  3 
so that our matrix solution would be
[ a_{n} b_{n} c_{n} ]
 =  [ a_{n1} b_{n1} c_{n1} ] 
* 
 1  2  2  
2  1  2 
2  2  3 


where a_{0} = 3, b_{0} = 4, c_{0} = 5.  (7) 
We can't help but notice the similarity in these matrices. In fact, in the case where a and b are consecutive integers, we used the triples where b = a + 1 . If we let a and b be consecutive integers in the order they appear when generated by equations (1), we would get the matrix equation

3  4  5  
21  20  29 
119  120  169 
 * 
 a_{11}  a_{12}  a_{13}  
a_{21}  a_{22}  a_{23} 
a_{31}  a_{32}  a_{33} 
 = 
 21  20  29  . 
119  120  169 
697  696  985 

The solution to this matrix equation yields the matrix
 1  2  2  . 
2  1  2 
2  2  3 
From this we see that all the matrices are the same except for making the elements in row one or two negative for the cases a and c consecutive odd integers or b and c consecutive integers, respectively.
Wait! There's More.
The matrices we found do more than we might have thought. The matrix we found to generate Pythagorean triples in which a and c are consecutive odd integers
 1  2  2  
2  1  2 
2  2  3 
actually does more. It, in fact, will preserve the difference c  a , regardless of what it is. If one were to multiply any Pythagorean triple, [ a b c ] with this matrix, we would get the triple
[ a + 2b + 2c 2a + b +2c 2a + 2b + 3c ]
and the difference between the first term and the last term is
(2a + 2b + 3c)  (a + 2b + 2c) = c  a.
So the difference is preserved. It should be noted that you will not get the next triple in line after [ a b c ], but you will get one for which c  a is the same.
The same is true for the other two cases. The matrix
 1  2  2  
2  1  2 
2  2  3 
will preserve the c  b . For if we multiply a triple [ a b c ] by this matrix we will get for the difference between the last term and the middle term
(2a  2b + 3c)  (2a  b + 2c) = c  b.
Finally, the matrix
 2  1  2  
1  2  2 
2  2  3 
will preserve the difference b  a . For, if we multiply a triple
[ a b c ] by this matrix we will get for the difference between the second term and the first term
(a + 2b + 2c)  (2a + b + 2c) = b  a.
A Final Note. We can use the matrix representations to get the following generators for the special Pythagorean triples.
For the case c = a + 2 , that is, a and c consecutive odd integers, we have
a_{n} = a_{n1} + 2b_{n1} + 4 
b_{n} = b_{n1} + 4 
c_{n} = a_{n1} + 2b_{n1} + 6 = a_{n} + 2 
where [ a_{n1} b_{n1} c_{n1} ] is a Pythagorean triple with c_{n1} = a_{n1} + 2 .
For the case c = b + 1 , that is, b and c consecutive integers, we have
a_{n} = a_{n1} + 2 
b_{n} = 2a_{n1} b_{n1} + 2 
c_{n} = 2a_{n1} + b_{n1} + 3 = b_{n} + 1 
where [ a_{n1} b_{n1} c_{n1} ] is a Pythagorean triple with c_{n1} = b_{n1} + 1 .
Finally, for the case b = a + 1 , that is, a and b consecutive integers, we have
a_{n} = 3a_{n1} + 2c_{n1} + 1 
b_{n} = 3a_{n1} + 2c_{n1} + 2 = a_{n} + 1 
c_{n} = 4a_{n1} + 3c_{n1} + 2 
where [ a_{n1} b_{n1} c_{n1} ] is a Pythagorean triple with b_{n1} = a_{n1} + 1 .
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