How Many Pythagorean Triples Where ( c leq 100 ) Can be Generated from the Fibonacci Method?

Introduction

In 1225, the renowned mathematician Leonardo of Pisa, known as Fibonacci, published Liber Quadratorum, or The Book of Squares. This work largely focused on quadratic equations and, as part of it, introduced a fascinating method for generating Pythagorean triples using sums of consecutive odd numbers.

Fibonacci's Discovery

Fibonacci noticed a remarkable relationship between sums of consecutive odd numbers and perfect squares. Specifically, the sum of the first ( n ) odd numbers equals ( n^2 ).

Mathematically:

[ sum_{k1}^n (2n-1) 1 3 5 ldots (2n-1) n^2 ]

Rephrasing the Discovery

When ( n ) is chosen such that the last summand is a perfect square, the equation can be rewritten. For instance, when ( n 1, 5, 13, 25, 41, 61, ldots ), the sums take the form:

Example 1:

[ 1 3 5 7 9 11 13 15 17 9^2 ]

Example 2:

[ (1 3 5 7 9) 12^2 13^2 ]

Here, the sum in parentheses is ( 5^2 ), leading to:

[ 5^2 12^2 13^2 ]

This directly results in the primitive Pythagorean triple {5, 12, 13}.

Generalizing to Odd Integers

For any odd integer ( k ), we can write ( k^2 2n - 1 ) and know that ( {k, n, n - 1} ) must be a primitive Pythagorean triple because ( n ) and ( n - 1 ) are coprime.

Application to ( c

Using this method, we can generate several Pythagorean triples where ( c a b c 3 4 5 5 12 13 7 24 25 9 40 41 11 60 61 13 84 85

Conclusion

Applying the Fibonacci method to generate Pythagorean triples is both elegant and insightful. By leveraging the properties of sums of consecutive odd numbers, one can easily construct a list of Pythagorean triples where ( c

Related Keywords

Pythagorean triples, Fibonacci method, c