The relation of Fibonacci numbers to the cubes $3^3+4^3+5^3=6^3$ and others?

Question

I. Cubes

This old post asks about generalizations to,

$$10^2+11^2+12^2=13^2+14^2$$

where all the terms are in arithmetic progression. We are familiar with,

$$3^3+4^3+5^3=6^3$$

Or by multiplying with a scaling factor $2^3$,

$$6^3+8^3+10^3=12^3$$

Label it as a $(\color{blue}{3,2,1})$, or $3$ cubes with a common difference of $2$ equal to $1$ cube. Compare to an equality I gave in the 2017 post,

$$47^3 +52^3 + 57^3 + 62^3 + 67^3 + 72^3 + 77^3 + 82^3 = 87^3 + 92^3 + 97^3$$

Or in equivalent form,

$$\sum_{k=0}^{7} (47+5k)^3 = \sum_{k=8}^{10} (47+5k)^3$$

so is a $(\color{blue}{8,5,3})$. Or,

$$\sum_{k=0}^{20} (322+13k)^3 = \sum_{k=21}^{28} (322+13k)^3$$

so is a $(\color{blue}{21,13,8})$. Rearrange to $(1,2,3,5,8,13,21)$ and we have the start of the Fibonacci sequence.

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II. Fibonacci numbers

It is easy enough to test the observation. The (sum of $m$ cubes starting with $a$) = (sum of $n$ cubes starting with $b$), with $m>n$ and common difference $d$,

$$\;a^3+(a+d)^3+(a+2d)^3+\cdots+\big(a+(m-1)d\big)^3=b^3+(b+d)^3+(b+2d)^3+\cdots+\big(b+(n-1)d\big)^3$$

is given by, after removing common factor $\frac14$,

$$m (2 a + d m - d)\,\big(2 a^2 + d (m - 1) (2 a + d m)\big) = n (2 b + d n - d)\, \big(2 b^2 + d (n - 1) (2 b + d n)\big)\;\tag1$$

with the condition that $\color{red}{b=a+md}$ since it is in arithmetic progression with the largest term of the LHS. If we label solutions to $(1)$ as $(n,d,m;a)$ with $m>n$, then,

$$(1,2,3;6)\\ (3,5,8;47)\\ (8,13,21;322)\\ (21,34,55;2190)\\ (55,89,144;14949)$$

so the Fibonacci pattern does continue.

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III. Question

Is it true that given the Fibonacci numbers $F_k$ with $F_1=F_2=1$ (and even index in blue),

$$F_k=1,\color{blue}1,2,\color{blue}3,5,\color{blue}8,13,\color{blue}{21},34,\color{blue}{55},\dots$$

then eq $(1)$ with $b=a+md\,$ has solutions $(n,d,m;a_k) = (F_{2k},\,F_{2k+1},\,F_{2k+2};\,a_k)$ where $a_k = 6, 47, 322, 2190, 14949,\dots$ is a yet to be determined integer sequence?

P.S. The $a_k$ is not yet in the OEIS.

Answer 1

Note: For the sequence $a_k$, thanks to Robert Israel for the generating function, Thomas Andrews for the recursion, and Michael Somos for the closed-form, crucial pieces of the puzzle. Andrews found that his recursion implied the divisibility property,

$$\frac{-a_k+a_{k+5}}{11}=-a_{k+1}+3\,a_{k+2}-3\,a_{k+3}+a_{k+4}$$

starting with the first six $a_k = (6,47,322,2190,14949,102284,\dots)$ so,

$$-a_1+a_6 = -6+102284 \equiv 0\, (\text{mod}\, 11)$$

etc. Somos found,

$$a_k = \frac{F_{4k}/F_{2k} - 2 + 3F_{4k} + F_{2k} + 13F_{2k}^2}4$$

And since we assume consecutive Fibonacci numbers,

$$(n,d,m) = (F_{2k},\;F_{2k+1},\;F_{2k+2})$$

so $n+d=m$. Substituting $(a,m)$ into (eq.1) with the given condition $b=a+md\,$ yields a complicated expression in three variables,

$$(F_{2k},\;F_{2k+1},\;F_{4k})$$

Since Fibonacci numbers have lots of relations, there must be a way to simplify this. We use variants of known identities,

$$F_{2k}=F_{2k+1}-F_{2k-1}$$ $$F_{4k}=F_{2k+1}^2-F_{2k-1}^2$$

to reduce the complicated expression into just two variables. I prefer to retain the consecutive $(x,y)=(F_{2k},\,F_{2k+1})$. Doing the substitution, the complicated expression finally factors into far simpler ones,

$$(x^2 + x y - y^2 + 1)\, (5 x^2 + 4 x y + y^2 - 1)\, (5 x^2 + 6 x y + 2 y^2 - 1)=0$$

with the first quadratic implying,

$$F_{2k}^2 + F_{2k}\,F_{2k+1} - F_{2k+1}^2 = -1$$

which is either a known identity or can be derived from known ones. The LHS factors over $\sqrt{5}$ while the LHS of the two other quadratics factor over $\sqrt{-1}$ and can be disregarded.

The three Fibonacci relations above can also simplify Somos' closed-form for the $a_k$ by eliminating $F_{4k}$. Given consecutive Fibonaccis $(x,y,z)=(F_{2k},\,F_{2k+1},\,F_{2k+2})$ yields two alternative closed-forms,

$$a_k = \frac{6x^2+(4x+1)y-y^2}2 = \frac{x-xy-yz+6xz}2$$

The equality of the two alternative formulas can be established by subtracting one from the other and they have the factor $(x+y-z)=0$ which Fibonacci numbers obey. And the substitution $(x,y)=(p-q,\,2q)$ on the first quadratic $x^2 + x y - y^2 = -1$ yields the Pell equation,

$$p^2-5q^2=-1$$

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Update (Jan 23, 2025): Courtesy of Adam Bailey's observation in the comments, we can look at the relationship between the initial point, midpoint, and endpoint of the equations, label them $(\alpha,\beta,\gamma)$. Using the same $(x,y,z)$ above, then they have the symmetric formulas,

\begin{align} \alpha_k &= \frac{\;x-xy-yz+6xz}2\\[4pt] \beta_k & = \,\frac{\alpha_k\,+\,\gamma_k\,}2 = 3xz\\[4pt] \gamma_k &= \frac{-x+xy+yz+6xz}2 \end{align}

with sequences,

$$\alpha_k = 6, 47, 322, 2190,\dots\\ \beta_k = 9, 72, 504, 3465,\dots\\ \gamma_k = 12, 97, 686, 4740\dots$$

and Bailey noticing that fractional multiples of $\beta_k$,

$\beta_k/3 = xz\, =\, F_{2k}\, F_{2k+2}\, = 3, 24, 168, 1155,\dots$ (A058038)

$\beta_k/9 = \dfrac{xz}3 = \dfrac{F_{2k}\, F_{2k+2}}3= 1,\, 8,\, 56,\, 385,\dots\,$ (A092521)

are in the OEIS.

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In summary, we have,

$$\sum_{k=1}^m a_k^3 = \sum_{k=1}^n b_k^3\tag1$$

or a sum of $m$ cubes equal to a sum of $n$ cubes in arithmetic progression $d$ such that $(n,d,m)$ are consecutive Fibonacci numbers, the smallest example being,

$$6^3+8^3+10^3=12^3\tag2$$

which is $(n,d,m)=(1,2,3)$ followed by,

$$\sum_{k=0}^{7} (47+5k)^3 = \sum_{k=8}^{10} (47+5k)^3\tag3$$

which is $(n,d,m)=(3,5,8)$. And so on.

P.S. While the math checks out, it seems hard to predict from first principles why Fibonacci numbers suddenly appear as a solution to equation $(1)$. Equality $(2)$ is well-known, and I've known about $(3)$ since 2017, but it's only now their connection to Fibonacci numbers was serendipitously noticed.

Answer 2

It looks to Maple like the sequence of $a$ values has generating function

$$ g(z) = \frac{z \left(z^{3}+3 z^{2}-19 z +6\right)}{\left(1-z\right) \left(z^{2}-7 z +1\right) \left(z^{2}-3 z +1\right)}$$

I have verified this for $k$ from $0$ to $1000$.