# 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.

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.

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.

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

• Thank you! It was reasonable to expect it to be a consequence of something much more general. How do you prove that any number $n \ge k +14$ can be expressed as the sum of $k$ nonzero squares for $k \ge 6$? Commented Mar 22, 2022 at 19:05
• @trilobita Conway gives the numbers not four nonzero squares. There are infinitely many but they are sparse. For $n > 50,$ say, either $n-1$ or $n-4$ or $n-9$ is the sum of four nonzero squares, hence $n$ itself is five. Next, if $n \geq n_k$ means it is the sum of $k$ nonzero squares, then $n \geq 1 + n_k$ means it is the sum of $1+k$ nonzero squares. Commented Mar 22, 2022 at 19:11
• @trilobita the book is available at maths.ed.ac.uk/~v1ranick/papers/conwaysens.pdf while the result used is on page 140, just called Theorem. I suspect the lists for four and five squares would be on the OEIS, let me check Commented Mar 22, 2022 at 19:17
• Thank you very much! Commented Mar 22, 2022 at 19:48

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 $$(2k+1)$$.

• I don't get it. Could you please write out your representations of $25$ as the sum of $6$ nonzero squares, then $169$ as the sum of $14$ nonzero squares? Commented Mar 22, 2022 at 18:24
• There is something which does not convince me: your last formula would imply that $\sum_{i=1}^{2k} F_i^2+1=0$. Also, $25=9+9+4+1+1+1$, and I think this is the only possibility Commented Mar 22, 2022 at 18:48
• Also, the number of terms on the RHS of your last formula is $(2k+1)+1$, not $F_{2k+1}+1$ Commented Mar 22, 2022 at 18:56