Fibonacci's Quickies

Fibonacci numbers Fn are formed recursively

 F0= 0
 F1= 1
 Fn= Fn-1 + Fn-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 (Fn, Fn+1, Fn+2), (Fn+3, Fn+4, Fn+5), (Fn+6, Fn+7, Fn+8), (Fn+9, Fn+10, Fn+11), where Fn is the nth Fibonacci number in the sequence 0, 1, 1, 2, 3, ...



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

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
  10. Fibonacci Numbers in Equilateral Triangle
  11. Binet's Formula by Inducion
  12. Binet's Formula via Generating Functions
  13. Generating Functions from Recurrences
  14. Cassini's Identity
  15. Fibonacci Idendtities with Matrices
  16. GCD of Fibonacci Numbers
  17. Binet's Formula with Cosines
  18. Lame's Theorem - First Application of Fibonacci Numbers

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Copyright © 1996-2017 Alexander Bogomolny

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 Fm, Fn, and Fp are the sides of a triangle. Without loss of generality, assume m < n < p. Then

 Fm + Fn≤ Fn-1 + Fn
  = Fn+1
  ≤ Fp,

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:

 Fm + Fn> Fp.

A contradiction.

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Copyright © 1996-2017 Alexander Bogomolny

There is no tetrahedron with the indicated vertices. The solution covers a more general problem: there is no tetrahedron with vertices (Fm, Fm+1, Fm+2), (Fn, Fn+1, Fn+2), (Fp, Fp+1, Fp+2), (Fq, Fq+1, Fq+2), for any four 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.

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Copyright © 1996-2017 Alexander Bogomolny


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