## Numbers: Their Tales, Types, and Treasures.

## Chapter 9: Number Relationships

### 9.5.FIBONACCI'S METHOD FOR FINDING PYTHAGOREAN TRIPLES

Fibonacci was perhaps one of the most influential mathematicians of the thirteenth century (__chapter 6__, __section 1__). In 1225, he published a book titled *Liber quadratorum* (*Book of Squares*), in which he demonstrated another relationship of numbers and stated the following:

I thought about the origin of all square numbers and discovered that they arise out of the increasing sequence of odd numbers; for the unity is a square and from it is made the first square, namely 1; that to this unity is added 3, making a second square, namely 4, with root 2; if to the sum is added the third odd number, namely 5, the third square is created, namely 9, with root 3; and thus sums of consecutive odd numbers and a sequence of squares always arise together in order.^{1}

Fibonacci was essentially describing the relationship that we discussed in __chapter 4__, __section 4__, that the sum of the first *n* odd numbers equals *n*^{2}:

1 + 3 + 5 +…+ (2*n* – 1) = *n*^{2}.

We visualize this statement again in __figure 9.3__. Notice how the squares beginning with the single square at the lower left increase in area by the consecutive odd numbers analogous to what we just established algebraically. This is a geometric analogue of this algebraic statement.

**Figure 9.3**

Fibonacci knew of the Pythagorean theorem and therefore was aware of Pythagorean triples—after all, he lived about 1,700 years after Pythagoras. He was able to generate these triples in the following way. Let's consider the sum of the first five odd natural numbers (as shown in __figure 9.3__), 1 + 3 + 5 + 7 + 9 = 5^{2}, whose last term (9) is a square number. The sum in the parentheses: (1 + 3 + 5 + 7) + 9 = 5^{2} is 16, which is also a square number because it is the sum of the first four odd numbers. So this equation can be rewritten as 16 + 9 = 25, which, surprisingly, gives us the primitive Pythagorean triple (3, 4, 5).

Let's consider another series of consecutive odd integers—one that ends with a square number—to convince ourselves that this scheme can really generate other primitive Pythagorean triples: Consider the sum of the first odd numbers up to the thirteenth, which happens to be a square number, 25 = 5^{2}:

(1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23) + 25 = 169 = 13^{2}.

Using the same procedure as above, we add all the terms prior to the last one—within the parentheses. Because it is the sum of the first twelve odd numbers, it must be equal to the square number 12^{2} = 144. Therefore, we obtain 144 + 25 = 169, or 12^{2} + 5^{2} =13^{2}. This gives us another primitive Pythagorean triple (5, 12, 13).

In general terms, Fibonacci's construction of Pythagorean triples can be described as follows: Choose any odd number *a* > 1. Every odd number has an odd square, therefore *a*^{2} is also an odd number, and we can write it in the form *a*^{2} = 2*n* + 1 for some natural number *n*. For example, we obtain *a*^{2} = 9 for *n* = 4, which is the fifth odd number, and *a*^{2} = 25 (for *n* = 12) is the thirteenth odd number. In general, 2*n* + 1 is the (*n* + 1)st odd number. Clearly, the sum of the first *n* + 1 odd numbers up to *a*^{2} = 2*n* + 1 is equal to (*n* + 1)^{2}:

1 + 3 + 5 +…+ (*a*^{2} – 2) + *a*^{2} = (*n* + 1)^{2}.

Here the summand (*a*^{2} – 2) is the *n*^{th} odd number, and the sum of the odd numbers up to (*a*^{2} – 2) is equal to *n*^{2}:

1 + 3 + 5 +…+ (*a*^{2} – 2) = *n*^{2}.

From this we obtain the following:

*n*^{2} + *a*^{2} = (*n* + 1)^{2}.

To summarize, for any odd number *a*, we can write *a*^{2} = 2*n* + 1, and the triple (*a*, *n*, *n* + 1) is a Pythagorean triple. Curiously, it is even a primitive Pythagorean triple because the only common factor of *n* and *n* + 1 is 1. From *a*^{2} = 2*n* + 1 we find and . We see that Fibonacci's method determines, for every odd number *a =* 3, 5, 7…, the primitive Pythagorean triple *a*, *b* = , *c* = . We can, therefore, conclude that there are an infinite number of primitive Pythagorean triples because there are infinitely many odd numbers. A few of these triples are listed in __table 9.1__.

We can further evaluate the numbers *b*_{n} in __table 9.1__ as

Therefore, Fibonacci's result means that

(*a** _{n}* = 2

*n*+ 1,

*b*

*= 2*

_{n}*n*(

*n*+ 1),

*c*

*= 2*

_{n}*n*(

*n*+ 1) + 1)

is a primitive Pythagorean triple for all natural numbers *n*.

We also note that *b** _{n}* = 4

*T*

*, where is the*

_{n}*n*th triangular number defined in

__chapter 4__,

__section 5__. The sequence of triangular numbers starts with 1, 3, 6, 10, 15, 21, 28, 36…, and the

*b*-values of

__table 9.1__are just four times these numbers. Therefore, Fibonacci's triples can also be written as (2

*k*+ 1, 4

*T*

*, 4*

_{k}*T*

*+ 1).*

_{k}**Table 9.1: Primitive Pythagorean triples obtained using Fibonacci's method.**

__Figure 9.4__ shows the right triangles corresponding to the first eight Pythagorean triples in __table 9.1__. Because the hypotenuse differs from the longer leg by 1, these triangles tend to become rather long and extended.

**Figure 9.4: Triangles corresponding to the Pythagorean triples in **__table 9.1__**.**