# Fibonacci's Quickies

Fibonacci numbers F_{n} are formed recursively

F_{0} | = 0 | |

F_{1} | = 1 | |

F_{n} | = F_{n-1} + F_{n-2}, n > 1 |

which results in the familiar sequence:

0, 1, 1, 2, 3, 5, 8, 13, ... |

This is one of the most remarkable sequences in mathematics which pops up uncannily in most unexpected circumstances. Fibonacci numbers deserve more than a fleeting and supercilious reference. Some are listed below. But here I mean to ask just two simple questions to answer which one only needs the understanding of rudimentary analytic geometry in two and three dimensions and an unadulterated definition of the Fibonacci sequence not loaded with derivation of the multitude of its numerous properties.

### Problem 1

[Trigg, #216]: Find Pythagorean triples whose sides are Fibonacci numbers.

### Problem 2

[Trigg, #209]: Find the volume of the tetrahedron with vertices _{n}, F_{n+1}, F_{n+2}),_{n+3}, F_{n+4}, F_{n+5}),_{n+6}, F_{n+7}, F_{n+8}),_{n+9}, F_{n+10}, F_{n+11}),_{n} is the nth Fibonacci number in the sequence 0, 1, 1, 2, 3, ...

### Reference

- C. W. Trigg,
*Mathematical Quickies*, Dover, 1985

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Copyright © 1996-2018 Alexander BogomolnyFind Pythagorean triples whose sides are Fibonacci numbers.

There are no right triangle whose sides are Fibonacci numbers. In fact mentioning the Pythagorean triples and implying the right triangles clouds the problem. There are no triangles of any kind with sides the Fibonacci numbers. The proof is by contradiction. Assume that F_{m}, F_{n}, and F_{p} are the sides of a triangle. Without loss of generality, assume

F_{m} + F_{n} | ≤ F_{n-1} + F_{n} | |

= F_{n+1} | ||

≤ F_{p}, |

with an equality only when m = n - 1 and p = n + 1. But for the three numbers to be the sides of a triangle it is necessary that they satisfy the triangle inequality:

F_{m} + F_{n} | > F_{p}. |

A contradiction.

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Copyright © 1996-2018 Alexander BogomolnyThere is no tetrahedron with the indicated vertices. The solution covers a more general problem: there is no tetrahedron with vertices (F_{m}, F_{m+1}, F_{m+2}), (F_{n}, F_{n+1}, F_{n+2}), (F_{p}, F_{p+1}, F_{p+2}), (F_{q}, F_{q+1}, F_{q+2}), for any four integers m, n, p, q. The reason for this is that any such vertex lies in the plane

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Copyright © 1996-2018 Alexander Bogomolny