# Integrality of a quotient of Fermat numbers

I try to prove that for every positive integers $$m\ge n$$, the following product is an integer: $$\prod_{k=0}^{n-1}\frac{2^{2^m}-2^{2^k}}{2^{2^n}-2^{2^k}}.$$

But no luck.

• @SamHopkins I must confess that, at first, I also didn't read the "is an integer" part of your question. XD Apr 14 at 0:16

## 1 Answer

The fraction equals $$\prod_{k=0}^{n-1}\frac{2^{2^m-2^k}-1}{2^{2^n-2^k}-1},$$ hence it suffices to show that $$\prod_{k=0}^{n-1}\frac{x^{2^m-2^k}-1}{x^{2^n-2^k}-1}\in\mathbb{Z}[x].$$ The numerator and the denominator factor into cyclotomic polynomials. If $$a(d)$$ and $$b(d)$$ are the multiplicties of the $$d$$-th cyclotomic polynomial in the numerator and the denominator, then it suffices to show that $$a(d)\geq b(d)$$. Now $$a(d)$$ equals the number of $$k\in\{0,\dotsc,n-1\}$$ such that $$d\mid 2^m-2^k$$, while $$b(d)$$ equals the number of $$k\in\{0,\dotsc,n-1\}$$ such that $$d\mid 2^n-2^k$$.

Let us write $$d=2^r e$$ with $$e$$ odd. If $$r\geq n$$, then $$a(d)=b(d)=0$$, and we are done. Assume now that $$r. Then $$a(d)$$ is the number of $$k\in\{r,\dotsc,n-1\}$$ such that $$e\mid 2^{m-k}-1$$, while $$b(d)$$ is the number of $$k\in\{r,\dotsc,n-1\}$$ such that $$e\mid 2^{n-k}-1$$. Let $$s$$ be the multiplicative order of $$2$$ modulo $$e$$. Then $$a(d)$$ is the number of elements of $$\{m-n+1,\dotsc,m-r\}$$ divisible by $$s$$, while $$b(d)$$ is the number of elements of $$\{1,\dotsc,n-r\}$$ divisible by $$s$$. Hence $$a(d)=\left\lfloor\frac{m-r}{s}\right\rfloor-\left\lfloor\frac{m-n}{s}\right\rfloor,\qquad b(d)=\left\lfloor\frac{n-r}{s}\right\rfloor,$$ and we are left with proving $$\left\lfloor\frac{m-r}{s}\right\rfloor\geq\left\lfloor\frac{m-n}{s}\right\rfloor+\left\lfloor\frac{n-r}{s}\right\rfloor.$$ However, this is clear from the inequality $$\lfloor u+v\rfloor\geq\lfloor u\rfloor+\lfloor v\rfloor$$, and we are done.

• Thanks a lot for this very clear answer. Apr 14 at 2:41
• @joaopa Thanks for the nice question. It was a pleasure answering it. Apr 14 at 2:44
• Sice note: since $\lfloor u+v\rfloor \le \lfloor u\rfloor + \lfloor v\rfloor + 1$, it seems this proof shows that the rational-function-that's-actually-a-polynomial is in fact squarefree as well. Apr 14 at 21:11
• @GregMartin Excellent point! Apr 14 at 21:20