If $n>1$, the square of the odd Fibonacci number $F(2n+1)$ can be written as the sum of exactly $F(2n+1)+1$ nonzero squares.

Question

While reading a paper by Owens (arXiv:1906.05913) about embeddings of rational homology balls in the complex projective plane, I found out the following somewhat unexpected number theory corollary (here $F(n)$ denotes the $n$-th Fibonacci number, with $F(1)=1$ and $F(2)=1$).

If $n>1$, the square of the odd Fibonacci number $F(2n+1)$ can be written as the sum of exactly $F(2n+1)+1$ nonzero squares.

I will put an answer with the proof. However, I am curious about whether a proof of this result can be obtained by purely numer-theoretic methods.

Answer 1

It follows from a result in Conway's little book The Sensual Quadratic Form, and a bit of computation, that every number $n \geq 34$ is the sum of five nonzero squares. That missing number $33$ is a one-off. There is a proof in Niven and Zuckerman where $34$ is replaced by the easier $170,$ pages 318-319 in the Fifth edition (with Montgomery, 1991).

With $k \geq 6:$ any number $n \geq k + 14$ can be expressed as the sum of $k$ nonzero squares.

The numbers that are not the sum of five nonzero squares are: $ 1,2, 3, 4, 6, 7, 9, 10, 12, 15, 18, 33 $ This list is at OEIS with references

not six: $1, 2, 3, 4, 5, 7, 8, 10, 11, 13, 16, 19.$

The induction step: if $n \geq n_k$ means that $n$ is the sum of $k$ nonzero squares, then $ n \geq 1 + n_k$ means that $n$ is the sum of $1+k$ nonzero squares

Answer 2

Owens proves that if $n>1$ the rational homology ball $B_{F(2n+1),F(2n-1)}$ can be embedded smoothly but not symplectically in $\mathbb{CP}^2$. From Donaldson's Theorem, Owens obtains an obstruction to the existence of such embeddings (Proposition 3.2). In particular, the intersection lattice $\Lambda_M$ of the complement of $B_{F(2n+1),F(2n-1)}$ in $\mathbb{CP}^2$, which has rank $1$ by Lemma 3.1, must embed in $\mathbb{Z}^{F(2n+1)+1}$, and its image must intersect nontrivially each unit vector. Since a generator of $\Lambda_M$ has self-pairing $F(2n+1)^2$, the conclusion follows.

Answer 3

Comment:This is just for some information. It probably can be used for an analytic answer to question.

We know :

$$\Sigma^{n}_{i=1} F_i^2=F_n\times F_{n+1}$$

Also:

$$F_n^2=F_{n-1}\times F_{k+1}+(-1)^{n-1}$$

Multiplying both sides by $F_n$ we obtain:

$$F_n^2\times F_n=F_{n-1}\times(F_n\cdot F_{(n+1)})+(-1)^{n-1}$$

Dividing both sides by $F_{n-1}$ we get:

$$F_n\times \Phi=\Sigma^n_{i=1} F_I^2+(-1)^{n-1}\times \Phi$$

If $n=2k+1$ then:

$$\Phi\big(F^2_{(2k+1)}-1\big)=\Sigma^{2k+1}_{i=1} F_i^2$$

The number of terms on RHS is $(2k+1)$.