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Pythagorean Triples, a2 + b2 = c2 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 = r2 - s2, b = 2rs , c = r2 + 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 | |
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| 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 | |
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| 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 = 4n2 - 1, b = 4n, c = 4n2 + 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 r2 + s2 = 2rs + 1 . Thus, r2 + s2 - 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
 | a11 | a12 | a13 | |
a21 | a22 | a23 |
a31 | a32 | a33 |
such that
 |
3 | 4 | 5 | |
20 | 21 | 29 |
119 | 120 | 169 |
| * |
 | a11 | a12 | a13 | |
a21 | a22 | a23 |
a31 | a32 | a33 |
| = |
 | 20 | 21 | 29 | . |
119 | 120 | 169 |
696 | 697 | 985 |
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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] |
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and
[20 21 29] |
* |
 | 2 | 1 | 2 | |
1 | 2 | 2 |
2 | 2 | 3 |
| = | | [119 120 169] |
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We therefore have a generator for the triples with a and b consecutive integers.
[ an bn cn ]
| = | [ an-1 bn-1 cn-1 ] |
* |
 | 2 | 1 | 2 | |
1 | 2 | 2 |
2 | 2 | 3 |
|
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where a0 = 3, b0 = 4, c0 = 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 |
| * |
 | a11 | a12 | a13 | |
a21 | a22 | a23 |
a31 | a32 | a33 |
| = |
 | 15 | 8 | 17 | . |
35 | 12 | 37 |
63 | 16 | 65 |
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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
[ an bn cn ]
| = | [ an-1 bn-1 cn-1 ] |
* |
 | -1 | -2 | -2 | |
2 | 1 | 2 |
2 | 2 | 3 |
|
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where a0 = 3, b0 = 4, c0 = 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 |
| * |
 | a11 | a12 | a13 | |
a21 | a22 | a23 |
a31 | a32 | a33 |
| = |
 | 5 | 12 | 13 | . |
7 | 24 | 25 |
9 | 40 | 41 |
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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
[ an bn cn ]
| = | [ an-1 bn-1 cn-1 ] |
* |
 | 1 | 2 | 2 | |
-2 | -1 | -2 |
2 | 2 | 3 |
|
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where a0 = 3, b0 = 4, c0 = 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 |
| * |
 | a11 | a12 | a13 | |
a21 | a22 | a23 |
a31 | a32 | a33 |
| = |
 | 21 | 20 | 29 | . |
119 | 120 | 169 |
697 | 696 | 985 |
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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
an = an-1 + 2bn-1 + 4 |
bn = bn-1 + 4 |
cn = an-1 + 2bn-1 + 6 = an + 2 |
where [ an-1 bn-1 cn-1 ] is a Pythagorean triple with cn-1 = an-1 + 2 .
For the case c = b + 1 , that is, b and c consecutive integers, we have
an = an-1 + 2 |
bn = 2an-1 bn-1 + 2 |
cn = 2an-1 + bn-1 + 3 = bn + 1 |
where [ an-1 bn-1 cn-1 ] is a Pythagorean triple with cn-1 = bn-1 + 1 .
Finally, for the case b = a + 1 , that is, a and b consecutive integers, we have
an = 3an-1 + 2cn-1 + 1 |
bn = 3an-1 + 2cn-1 + 2 = an + 1 |
cn = 4an-1 + 3cn-1 + 2 |
where [ an-1 bn-1 cn-1 ] is a Pythagorean triple with bn-1 = an-1 + 1 .
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