Fibonacci's Quickies

Fibonacci numbers F_n are formed recursively

which results in the familiar sequence:

This is one of the most remarkable sequences in mathematics which pops up uncannily in most unexpected circumstances. Fibonacci numbers65 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.

Solution

Problem 2

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

Solution

Reference

1. C. W. Trigg, Mathematical Quickies, Dover, 1985, #174

Fibonacci Numbers

1. Ceva's Theorem: A Matter of Appreciation
2. When the Counting Gets Tough, the Tough Count on Mathematics
3. I. Sharygin's Problem of Criminal Ministers
4. Single Pile Games
5. Take-Away Games
6. Number 8 Is Interesting
7. Curry's Paradox
8. A Problem in Checker-Jumping
9. Fibonacci's Quickies

Golden Ratio

1. Golden Ratio in Geometry
2. Golden Ratio in an Irregular Pentagon
3. Golden Ratio in a Irregular Pentagon II
4. Inflection Points of Fourth Degree Polynomials
5. Wythoff's Nim
6. Inscribing a regular pentagon in a circle – and proving it
7. Cosine of 36 degrees
8. Continued Fractions

Find 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 m < n < p. Then

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:

A contradiction.

There 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 three integers m, n, p, q. The reason for this is that any such vertex lies in the plane x + y = z, such that all four are coplanar. Even accepting that coplanar points may form a tetrahedron, the volume of the latter is bound to be zero.