# Texas Mathematics Teacher Spring/Summer 2018 - 28

```Pythagoras Unchained
To identify patterns, we looked at the first and second order differences in the sequences. Below is sequence 1.
Sequence 1:
1
2
5
10
17
26
First Order Difference:
1
3
5
7
9
Second Order Difference:
2
2
2
2
In sequence 2, the first and second order differences are identical to those for sequence 1:
Sequence 2:
4
5
8
13
20
29
First Order Difference:
1
3
5
7
9
Second Order Difference:
2
2
2
2
Likewise, in sequence 3:
Sequence 3:
9
10
13
18
25
34
First Order Difference:
1
3
5
7
9
Second Order Difference:
2
2
2
2
Also, sequence 4:
Sequence 4:
16
17
20
25
32
41
First Order Difference:
1
3
5
7
9
Second Order Difference:
2
2
2
2
When the second order difference of a sequence is constant, one knows the sequence can be produced by a
quadratic function. That is, there exist a, b, and c such that, in sequence 1, a ⋅12 + b ⋅1 + c =1 , a ⋅ 22 + b ⋅ 2 + c =,
2
a ⋅ 32 + b ⋅ 3 + c =,
5 and so on. Because there are 3 unknowns, 3 equations are sufficient to find the values of
a, b, and c.
This leads to the following system of equations for sequence 1:
1a + 1b + c =
1
4a + 2b + c =
2 which, written as an augmented matrix, is
9a + 3b + c =
5

The row reduced echelon form of this matrix is

1 1 1 1
4 2 1 2
9 3 1 5

1 0 0 1
0 1 0 −2
0 0 1 2

That is, a =
1, b =
−2, and c =
2 . Therefore, the function f1 ( n ) = n 2 − 2n + 2 will produce sequence 1. As n takes
on the values 1, 2, 3, 4, ..., f1 ( n ) takes on the values 1, 2, 5, 10, ... which are the radicands of the lengths of
segments in column 1.

28

| Spring/Summer 2018

Texas Mathematics Teacher

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http://www.brightcopy.net/allen/txmt/68-01
http://www.brightcopy.net/allen/txmt/67-01
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