# Texas Mathematics Teacher Spring/Summer 2018 - 29

```Pythagoras Unchained
Repeating the process results in the following equations for the remaining sequences:
Sequence 1 is produced by the function f1 ( n ) = n 2 − 2n + 2

Sequence 2 is produced by the function f 2 ( n ) = n 2 − 2n + 5 ;

Sequence 3 is produced by the function f3 ( n ) = n 2 − 2n + 10 ; and
Sequence 4 is produced by the function f 4 ( n ) = n 2 − 2n + 17

Notice the constant term c of f1 ( n ) is 2, the constant term of f 2 ( n ) is 5, the constant term of f3 ( n ) is 10,
and the constant term of f 4 ( n ) is 17. We have seen this sequence 2, 5, 10, 17, ... before. In general, it appears
the constant terms of the functions are the elements of sequence 1. However, the constant term of f1 ( n ) is the
second term of sequence 1, the constant term of f 2 ( n ) is the third term of sequence 1 and so on. We see the
value of c in any given f m is the f ( m + 1) term from sequence 1. To obtain a formula for the c value of equation
f m we must translate the function of f1 by 1 unit. The resulting equation is:
c fm = f1 ( m + 1) = (m + 1) 2 − 2(m + 1) + 2 or m 2 + 1 .

Recalling that the lengths of the segments that can be produced are the square roots of the elements of the
sequences, we see all the lengths which can be formed on an infinitely large piece of paper on which dots have
been placed in a square grid pattern 1 unit apart can be produced by the function:
Fm ( n ) =

n 2 − 2n + cm , where c=
m 2 + 1 , where m and n are positive integers.
m

Although this has been explored through patterns, it is important to connect this formula to the Pythagorean
theorem. Given a right triangle with legs of lengths a and b and hypotenuse of length c, we can see that
Fa ( b + 1=
) Fb ( a + 1=) c .
For example, given a right triangle with legs of lengths a= 3 and b= 4 and hypotenuse of c= 5.
Fa ( b + 1)= F3 ( 4 + 1)= F3 ( 5 )=

Since c3 = 32 + 1 = 10 , F3 ( 5 ) =

52 − 2(5) + c3
52 − 2(5) + 10 =

Similarly, Fb ( a + 1)= F4 ( 3 + 1)= F4 ( 4 )=

25 = 5

42 − 2(4) + c4 = 42 − 2(4) + 17 =

25 = 5

Figure 6. Example when solutions are not unique.
It directly follows then that although this formula gives all possible lengths,
solutions are not necessarily unique. For example, not only F3 ( 5 ) and F4 ( 4 )
give a result of 5, but F5 (1) does as well (see left Figure 6).
Applications to Teaching
The authors of the previously mentioned book aimed at creating problems structured to provide opportunities,
primarily for teachers, for mathematical discovery through exploration of patterns and through making
connections across different content domains within mathematics. Explorations such as this one are very much
in this spirit. The challenge for teachers is to create similar experiences and opportunities for their students that

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Spring/Summer 2018 | 29

```
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