How Many Geometric Polyhedra?

There are infinitely many solutions for edges $E = 2F_{2n}^2$ , faces $F = 2F_{2n - 1}^2$ , and vertices $V = 2F_{2n}F_{2n - 1}$ for all Fibonacci numbers $F_{2n}$ and $F_{2n - 1}$ for integers $n > 1$ .

These equations satisfy the geometric progression, because $V^2 = (2F_{2n}F_{2n - 1})^2 = 2F_{2n}^2 \cdot 2F_{2n - 1}^2 = EF$ .

These equations also satisfy Euler’s Formula $V + F - E = 2$ , because

$V + F - E$

$= 2F_{2n}F_{2n - 1} + 2F_{2n - 1}^2 - 2F_{2n}^2$

$= 2(F_{2n}F_{2n - 1} + F_{2n - 1}^2 - F_{2n}^2)$

$= 2(F_{2n - 1}(F_{2n} + F_{2n - 1}) - F_{2n}^2)$

$= 2(F_{2n - 1}F_{2n + 1} - F_{2n}^2)$ (since $F_{2n} + F_{2n - 1} = F_{2n + 1}$ )

$= -2(F_{2n}^2 - F_{2n - 1}F_{2n + 1})$

$= -2(-1)^{2n - 1}$ (by Cassini's identity)

$= 2$

Furthermore, the requirement that $2E \geq 3F$ is also satisfied because by substitution it becomes $2 \cdot 2F_{2n}^2 \geq 3 \cdot 2F_{2n}F_{2n - 1}$ which simplifies to $\frac{F_{2n}}{F_{2n-1}} \geq \frac{3}{2}$ which is true for all $n > 1$ since $\frac{F_{2n}}{F_{2n-1}}$ approaches $\phi = \frac{1 + \sqrt{5}}{2}$ . Also, $2E \geq 3F \geq 3V$ , so $2E \geq 3V$ .

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When $n = 2$ , $E = 18$ , $V = 12$ , and $F = 8$ , a hexagonal prism. Also of note is a hexagonal bipyramid when $V = 8$ , $F = 12$ , and $E = 18$ .

Polyhedra with $(F, V, E)$ values $(50, 80, 128)$ and $(338, 546, 882)$ (and so on) can also be derived with these equations.