Consider the Young diagram of a partition $\lambda = (\lambda\_1,\ldots,\lambda\_k)$. For a square $(i,j) \in \lambda$, define the *hook numbers* $h\_{(i,j)} = \lambda\_i + \lambda\_j' -i - j +1$ where $\lambda'$ is the *conjugate* of $\lambda$.


The [hook-length formula](https://en.wikipedia.org/wiki/Hook_length_formula) shows, in particular, that if $\lambda\vdash n$ then 
$$\text{$n!\prod\_{\square\,\in\,\lambda}\frac1{h\_{\square}}$} \qquad \text{is an integer}.$$
Recall the *Fibonacci numbers* $F(0)=0, \, F(1)=1$ with $F(n)=F(n-1)+F(n-2)$. Define $[0]!\_F=1$ and $[n]!\_F=F(1)\cdot F(2)\cdots F(n)$ for $n\geq1$.



> 
> **QUESTION.** Is it true that 
>  $$\text{$[n]!\_F\prod\_{\square\,\in\,\lambda}\frac1{F(h\_{\square})}$} \qquad \text{is an integer}?$$
> 
> 
> 



hook-length formula: "Fibonaccized" Part I
Sam is correct of course about $q$-hook formula. Below is a short self-contained proof not relying on such advanced combinatorics.


Denote $h\_1>\ldots>h\_k$ the set of hook lengths of the first column of diagram $\lambda$. Then the multiset of hooks is $\cup\_{i=1}^k \{1,2,\ldots,h\_i\}\setminus \{h\_i-h\_j:i<j\}$ and $n=\sum\_i h\_i-\frac{k(k-1)}2$. 


Recall that $F(m)=P\_m(\alpha,\beta)=\prod\_{d|m,d>1}\Phi\_d(\alpha,\beta)=\prod\_d (\Phi\_d(\alpha,\beta))^{\eta\_d(m)}$, where 


$\alpha,\beta=(1\pm \sqrt{5})/2$; 


$P\_n(x,y)=x^{n-1}+x^{n-2}y+\ldots+y^{n-1}$; 


$\Phi\_d$ are homogeneous cyclotomic polynomials;


$\eta\_d(m)=\chi\_{\mathbb{Z}}(m/d)$ (i.e., it equals to 1 if $d$ divides $m$, and to 0 otherwise). 


Therefore it suffices to prove that for any fixed $d>1$ we have
$$
\sum\_{m=1}^n \eta\_d(m)+\sum\_{i<j}\eta\_d(h\_i-h\_j)-\sum\_{i=1}^k\sum\_{j=1}^{h\_i}\eta\_d(j)\geqslant 0.\quad (\ast)
$$
$(\ast)$ rewrites as
$$
[n/d]+|i<j:h\_i\equiv h\_j \pmod d|-\sum\_{i=1}^k [h\_i/d]\geqslant 0.\quad (\bullet)
$$
LHS of $(\bullet)$ does not change if we reduce all $h\_i$'s modulo $d$ (and accordingly change $n=\sum\_i h\_i-\frac{k(k-1)}2$, of course), so we may suppose that $0\leqslant h\_i\leqslant d-1$ for all $i$. For $a=0,1,\dots, d-1$ denote $t\_a=|i:h\_i=a|$. Then $(\bullet)$ rewrites as
$$
\left[\frac{-\binom{\sum\_{i=0}^{d-1} t\_i}2+\sum\_{i=0}^{d-1} it\_i}d\right]+
\sum\_{i=0}^{d-1} \binom{t\_i}2\geqslant 0. \quad (\star)
$$


It remains to observe that LHS of $(\star)$ equals to
$$
\left[\frac1d\sum\_{i<j}\binom{t\_i-t\_j}2 \right].
$$


