Fermat numbers question

Question

I am looking for direction in my question. The question is to prove, by long polynomial division, that: \begin{equation} 2^{2^{n+1}}-1\ |\ 2^{F_n-1}-1 \end{equation}

I have tried a lot of things, like calling:\begin{equation} t=2^{2^n} \end{equation} and trying long polynomial division right away, but none of this worked successfully.

It would improve your setup of the problem to include a definition of Fermat number $F_n$, so that the divisibility property you want to show is framed in a clearer way. — hardmath, Nov 24 '20 at 18:10
so I need to prove that, in order to use it with Fermat numbers? because in class I remember us proving that. — dfdf1, Nov 24 '20 at 18:29
Welcome to Mathematics Stack Exchange. Please use MathJax. Can't you use $2^a-1|2^b-1$ if $a|b$? — J. W. Tanner, Nov 27 '20 at 17:49
(Fn) - 1 ,and I am sorry I dont know how to use MathJax, I'll try to learn — gftyh, Nov 27 '20 at 18:06
look for this question you"ll understand what I mean :Fermat numbers question — gftyh, Nov 27 '20 at 18:59
they didnt show how they did the polynomial division in that question, and Im asking for it. — gftyh, Nov 27 '20 at 19:08
In this question I just used the fact that in order to prove your result you just need to show that $2^{2^{n+1}}-1 \mid (2^{2^{n+1}})^{2^k}-1$ for some non negative integer $k$. Calling $t=2^{2^{n+1}}$ you just need to show that the polynomial $t-1 \mid t^{2^k}-1$ using long division. — PAM1499, Nov 27 '20 at 20:28

PAM1499 · Accepted Answer · 2020-11-24T21:28:46.620

0

With the tip given by lulu you have $gcd(2^{2^{n+1}}-1,2^{F_n-1}-1)=2^{gcd(2^{n+1},F_n-1)}-1=2^{gcd(2^{n+1},2^{2^{n}})}-1$. Notice that for all $n \in \mathbb{N}$ $n+1\leq 2^n$(use induction to see this). Hence $2^{n+1} \mid 2^{2^n}$ and $gcd(2^{n+1},2^{2^{n}})=2^{n+1}$. So, $gcd(2^{2^{n+1}}-1,2^{F_n-1}-1)=2^{2^{n+1}}-1$ and we have that $2^{2^{n+1}}-1 \mid 2^{F_n-1}-1$.

Another try Notice (again) that $n+1\leq 2^n$. Hence there is some $k \in \mathbb{N}$ such that $(n+1)+k=2^n$. Then $2^{2^n}=2^{(n+1)+k}=2^{n+1} \cdot 2^k$. Notice then that $$ 2^{F_n-1}-1=2^{2^{2^n}}-1=2^{2^{n+1} \cdot 2^k}-1=(2^{2^{n+1}})^{2^k}-1. $$ So, to prove that $2^{2^{n+1}}-1 \mid 2^{F_n-1}-1$ is equivalent to prove that $2^{2^{n+1}}-1 \mid (2^{2^{n+1}})^{2^k}-1$. Now call $t=2^{2^{n+1}}$ and you need to prove that $t-1 \mid t^{2^k}-1$. Can you use long division to finish this?

edited Nov 24 '20 at 21:28

answered Nov 24 '20 at 19:19

PAM1499

1,177

But I need to do that via long polynomial division. – dfdf1 Nov 24 '20 at 20:29
Oh, I see. Since you said "in class I remember us proving that" I thought you were talking about this property. I will delete the answer then if you prefer – PAM1499 Nov 24 '20 at 20:33
Yes I prefer. But thank you on the time man!!! I would like to see, if u got a direction for me to solve this via long polynomial division. – dfdf1 Nov 24 '20 at 20:36
Actually, $t^{2^k}-1=(t-1)(t^{2^k-1}+t^{2^k-2} \cdots + t + 1)$. – PAM1499 Nov 24 '20 at 22:40
But yes,if you do this substitution on the start you get $2^{2^{2^n}}-1=2^t-1$ and this is not a polynomial. – PAM1499 Nov 24 '20 at 22:42
Oh yes yes. I forgot the 2, I am sorry. thank you!!! – dfdf1 Nov 24 '20 at 22:42
That's ok! Hope it helps ;) – PAM1499 Nov 24 '20 at 22:43
Yes I got it. thank you!!! – dfdf1 Nov 24 '20 at 22:43
Glad to help my friend! – PAM1499 Nov 24 '20 at 22:43

score 0 · Answer 2 · answered Nov 27 '20 at 21:47

If you want to see how to use Polynomial long division as described in here https://en.wikipedia.org/wiki/Polynomial_long_division to show that $t-1 \mid t^{2^k}-1$ let's do it.

Here the dividend is $t^{2^{k}}-1$ and the divisor is $t-1$. The first step of Polynomial long division is to divide the first term of the dividend ($t^{2^{k}}$) by the highest term of the divisor ($t$) $$ \frac{t^{2^{k}}}{t}=t^{2^{k}-1}. $$ Now we multiply the divisor ($t-1$) by the result just obtained ($t^{2^{k}-1}$) to get $$ t^{2^{k}-1}(t-1)=t^{2^{k}}-t^{2^{k}-1}. $$ Now subtract the product just obtained ($t^{2^{k}}-t^{2^{k}-1}$) from the appropriate terms of the original dividend ($t^{2^{k}}-1$) and we get $t^{2^{k}-1}-1$. Now we repeat the process with $t^{2^{k}-1}-1$ as the dividend. Repeating the process one gets $t^{(2^{k}-1)-1}-1=t^{2^{k}-2}-1$. And in the $r-1$-th step one gets $t^{2^{k}-r}-1$(prove it). Suppose WLOG that we must stop the process at this step. This means that the degree of the polynomial $t^{2^{k}-r}-1$ is smaller then the degree of the polynomial $t-1$. Since the degree of $t-1$ is $1$ we know that the degree of $t^{2^{k}-r}-1$ must be $0$. Hence we must have that $r=2^k$ and that $t^{2^{k}-r}-1=t^0-1=0$. So we have that the remainder of the long division is $0$. And we are done.

Fermat numbers question

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