Fibonacci and Pythagoras

It's easy to discover that $F_{2n-1} = {F_n}^2 + {F_{n-1}}^2$ , thus we have ${F_{2n-1}}^2 = {({F_n}^2)}^2 + {({F_{n-1}}^2)}^2 + 2{{F_n}^2}{{F_{n-1}}^2} \\ = {({F_n}^2 - {F_{n-1}}^2)}^2 + {(2{F_n}{F_{n-1}})}^2$

Since all terms here appear as squares of positive integers, the statement is proved.

(We can prove that $F_{2n-1} = {F_n}^2 + {F_{n-1}}^2$ in this way:

$F_{2n-1} = F_{2n-3} + F_{2n-2} \\ = {F_1}{F_{2n-3}} + {F_2}{F_{2n-2}} \\ = {F_1}{F_{2n-3}} + {F_2}{({F_{2n-3}} + {F_{2n-4}})}\\ = {(F_1 + F_2)}{F_{2n-3}} + {F_2}{F_{2n-4}} \\ = {F_2}{F_{2n-4}} + {F_3}{F_{2n-3}}\\ = {F_3}{F_{2n-5}} + {F_4}{F_{2n-4}}\\ = …\\ = {F_{n-1}}^2 + {F_n}^2 .$

Or simply take the closed-form formula (Binet's formula) and verify it.)

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Another way to prove that $F_{2n-1} = {F_n}^2 + {F_{n-1}}^2$ using induction. Less elegant than the method mentioned above, but maybe easier to think of and more natural.

Suppose that $F_{2n-1} = {F_n}^2 + {F_{n-1}}^2$ holds for all $n ≤ k$ . Thus we have $F_{2k-1} = {F_k}^2 + {F_{k-1}}^2$ and $F_{2k-3} = {F_{k-1}}^2 + {F_{k-2}}^2$ . It is to show that the statement also holds for $n = k+1$ , that is, $F_{2k+1} = {F_{k+1}}^2 + {F_k}^2$ .

The first step is to find the relation between $F_{2k+1}$ , $F_{2k-1}$ , and $F_{2k-3}$ .

$F_{2k+1} = F_{2k} + F_{2k-1}\\ = 2 F_{2k-1} + F_{2k-2}\\ = 3 F_{2k-2} + 2 F_{2k-3}$

From the two equations above we can eliminate $F_{2k-2}$ , and get

$F_{2k+1} = 3 F_{2k-1} - F_{2k-3}\\ = 3({F_k}^2 + {F_{k-1}}^2) - ({F_{k-1}}^2 + {F_{k-2}}^2)\\ = 3{F_k}^2 + 2{F_{k-1}}^2 - {F_{k-2}}^2\\ = 3{F_k}^2 + 2{F_{k-1}}^2 - (F_k - F_{k-1})^2\\ = 2{F_k}^2 + {F_{k-1}}^2 + 2{F_k}{F_{k-1}}\\ = (F_k + F_{k-1})^2 + {F_k}^2\\ = {F_{k+1}}^2 + {F_k}^2 .$

And so we have our statement proved.

Applying Binet's Formula, we have $F_n \; = \; \tfrac{1}{\sqrt{5}}\big[\phi^n - \psi^n\big] \hspace{2cm} n \ge 0$ where $\phi \; = \; \tfrac12\big(1 + \sqrt{5}\big) \hspace{2cm} \psi \; = \; \tfrac12\big(1 - \sqrt{5}\big)$ so that $\phi + \psi = 1$ , $\phi\psi = -1$ , and $\phi^2 - \phi - 1 = \psi^2 - \psi-1 = 0$ . Then $\begin{aligned} 2F_nF_{n-1} & = \; \tfrac25\big(\phi^{2n+1} + \psi^{2n+1} - \phi^n\psi^{n+1} - \phi^{n+1}\psi^n\big) \; = \; \tfrac25\big[\phi^{2n+1} + \psi^{2n+1} + (-1)^{n+1}\big] \\ F_{n-1}F_{n+2} & = \; \tfrac15\big(\phi^{2n+1} + \psi^{2n+1} - \phi^{n-1}\psi^{n+2} - \phi^{n+2}\psi^{n-1}\big] \; = \; \tfrac15\big[\phi^{2n+1} + \psi^{2n+1} + (-1)^n(\phi^3 + \psi^3)\big] \; = \; \tfrac15\big[\phi^{2n+1} + \psi^{2n+1} + 4(-1)^n\big] \end{aligned}$ so that $\begin{aligned} \big(2F_nF_{n+1}\big)^2 + \big(F_{n-1}F_{n+2}\big)^2 & = \; \tfrac{1}{25}\left[ \begin{array}{l} 4(\phi^{2n+1}+\psi^{2n+1})^2 + 8(-1)^{n+1}(\phi^{2n+1} + \psi^{2n+1}) + 4 \\ + (\phi^{2n+1} + \psi^{2n+1})^2 + 8(-1)^n(\phi^{2n+1} + \psi^{2n+1}) + 16\end{array}\right] \\ & = \; \tfrac15\big[(\phi^{2n+1} + \psi^{2n+1}\big)^2 + 4\big] \; = \; \tfrac15\big[\phi^{2n+1} - \psi^{2n+1}\big]^2 \end{aligned}$ and hence $\big(2F_nF_{n+1}\big)^2 + \big(F_{n-1}F_{n+2}\big)^2 \; = \; F_{2n+1}^2$ for all $n \ge 1$ , and the two terms $2F_nF_{n+1}$ and $F_{n-1}F_{n+2}$ are nonzero for $n \ge 2$ , and hence $F_{2n+1}$ is the hypotenuse of a Pythagorean triad for all $n \ge 2$ , making the answer $\boxed{\mathrm{Yes}}$ .

Any sum of two squares $p^2 + q^2$ is the hypotenuse in a Pythagorean triple: $(a,b,c) = (p^2-q^2, 2pq, p^2+q^2)$ for some integers $p,q$ .

Since $F_{2n-1} = F_n^2 + F_{n-1}^2$ , we simple set $p = F_n$ and $q = F_{n+1}$ to obtain the desired conclusion.

We can derive it directly using the formula. Most elegant way is to use linear algebra.

I love the proofs. I solved the problem by observation. Unfortunately I didn’t notice that F(2n-1) = Fn^2 + F(n-1)^2. I saw a recursion that produced all the Pythagorean triples. Given Pythagorean triples (a(n-2), b(n-2), F(n-2)) and (a(n), b(n) and F(n)), the next triple is given by a(n+2) = a(n-2) + F(n), b(n+2) = b(n)+F(n+1) - a(n-2), and, of course, F(n+2).