Pythagorean Triangle from Fibonacci Numbers
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Theorem
Take $4$ consecutive Fibonacci numbers:
- $F_n, F_{n + 1}, F_{n + 2}, F_{n + 3}$
Let:
- $a := F_n F_{n + 3}$
- $b := 2 F_{n + 1} F_{n + 2}$
- $c := F_{2 n + 3}$
Then:
- $a^2 + b^2 = c^2$
and:
- $\dfrac {a b} 2 = F_n \times F_{n + 1} \times F_{n + 2} \times F_{n + 3}$
That is, if the legs of a right triangle are the product of the outer terms and twice the inner terms, then:
- the hypotenuse is the Fibonacci number whose index is half the sum of the indices of the four given Fibonacci numbers.
- the area is the product of the four given Fibonacci numbers.
Proof
By definition of Fibonacci numbers:
- $F_n = F_{n + 2} - F_{n + 1}$
and:
- $F_{n + 3} = F_{n + 2} + F_{n + 1}$
Then $a$ can be expressed as:
\(\ds a\) | \(=\) | \(\ds \paren {F_{n + 2} - F_{n + 1} } \paren {F_{n + 2} + F_{n + 1} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds {F_{n + 2} }^2 - {F_{n + 1} }^2\) |
Now consider:
\(\ds \) | \(\) | \(\ds F_{n + 2} F_{n + 3} - F_n F_{n + 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds F_{n + 2} \paren {F_{n + 2} + F_{n + 1} } - \paren {F_{n + 2} - F_{n + 1} } F_{n + 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds {F_{n + 2} }^2 + {F_{n + 1} }^2\) | multiplying out |
and it can be seen that the ordered triple:
- $\paren {{F_{n + 2} }^2 - {F_{n + 1} }^2, 2 F_{n + 1} F_{n + 2}, {F_{n + 2} }^2 + {F_{n + 1} }^2}$
is in the form of Solutions of Pythagorean Equation.
It remains to be shown that:
- ${F_{n + 2} }^2 + {F_{n + 1} }^2 = F_{2 n + 3}$
We have:
\(\ds F_{2 n + 3}\) | \(=\) | \(\ds F_{\paren {n + 1} + \paren {n + 2} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds F_{\paren {n + 2} } F_{\paren {n + 1} + 1} + F_{\paren {n + 2} - 1} F_{\paren {n + 1} }\) | Honsberger's Identity | |||||||||||
\(\ds \) | \(=\) | \(\ds {F_{n + 2} }^2 + {F_{n + 1} }^2\) |
and the proof is complete.
$\Box$
The area is trivial:
\(\ds a\) | \(=\) | \(\ds F_n F_{n + 3}\) | ||||||||||||
\(\ds b\) | \(=\) | \(\ds 2 F_{n + 1} F_{n + 2}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {a b} 2\) | \(=\) | \(\ds \frac {F_n F_{n + 3} \times 2 F_{n + 1} F_{n + 2} } 2\) | |||||||||||
\(\ds \) | \(=\) | \(\ds F_n \times F_{n + 1} \times F_{n + 2} \times F_{n + 3}\) |
$\blacksquare$
Sources
- 1948: Charles W. Raine: Pythagorean Triangles from the Fibonacci Series 1, 1, 2, 3, 5, 8, (Scripta Math. Vol. 14: p. 164)
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $5$
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $5$