Linked Questions
13 questions linked to/from A theorem about prime divisors of generalized Fermat numbers?
16
votes
3
answers
806
views
Mental Primality Testing
At a trivia night, the following question was posed: "What is the smallest 5 digit prime?"
Teams (of 4) were given about a minute to write down their answer to the question. Obviously, the answer is ...
- 723
4
votes
4
answers
1k
views
Efficiently prove $2$ generates $(\mathbb{Z}/19\mathbb{Z})^*$ [Order Testing]
So I'm first asked to compute, mod 19, the powers of 2,
$$2^{2},2^{3},2^{6},2^{9}$$
which I compute as
$$4,8,7,18$$
respectively.
I'm then asked to prove that 2 generates $(\mathbb{Z}/19\...
- 5,562
8
votes
2
answers
3k
views
How to prove that if a prime $p$ divides a Fermat Number then $p=k\cdot 2^{n+2}+1$?
If a prime $p$ divides a Fermat Number then $p=k\cdot 2^{n+2}+1$?
Does anyone know a simple/elementary proof?
- 449
5
votes
4
answers
1k
views
Why are the first 5 Fermat numbers prime?
The $n$th Fermat number $F_n$ is defined as $F_n = 2^{2^n}+1$. The first five Fermat numbers, $F_0,F_1,F_2,F_3,F_4$, are all prime. Why is this?
It seems like a fairly surprising coincidence that ...
- 3,461
6
votes
2
answers
614
views
$F_{5}=2^{2^{5}}+1 $ is not prime
In 1637 Fermat stated that $F_{5}=2^{2^{5}}+1=4294967297 $ is prime. On the contrary Euler showed that:
$F_{5}=(2^{16})^2+1^{2}=62264^{2}+20499^2$.
Because of Fermat I know that $F_{5}$ mod $4 $ is $...
- 415
0
votes
1
answer
3k
views
Let $p$ be a prime divisor of the Fermat number $F_n = 2^{2^ n} + 1$. Prove that $p$ must have the form $2^{n+1}k + 1$. [duplicate]
Let $p$ be a prime divisor of the Fermat number $F_n = 2^{2^n} + 1$. Prove that $p$ must have the form $2^{n+1}k + 1$.
Thank you in advance.
- 687
12
votes
1
answer
252
views
Family of GCDs all equal to $2$
Is it true that
$$\gcd\left(5^{2^n} + 1, 13^{2^n} + 1\right) = 2$$
for all $n \in \mathbf{Z}_{\geq 0}$?
I'm continually stumped with this and verifying it numerically is quite expensive very quickly (...
- 235
3
votes
1
answer
404
views
There is no Fermat number that is divisible by $97$.
Because $97$ is a prime number of the form $3.2^n+1$, The order of $2$ modulo $97$ is either $3$, $2^k$ or $3\cdot2^k$ for some $0\le k\le n$.
Since $2^{2^k}-1=F_0F_1F_2\cdots F_{k-1}$,The order of $...
user118535
2
votes
1
answer
207
views
Divisor of $2^{2^n} + 1$ modulus $2^{n+1}$
I'm stuck at the following question: let $d$ divide $2^{2^n} +1 $. Show that $d \equiv 1 \pmod{2^{n+1}}$.
All I've managed to do so far was to notice that $2^{n+1} \mid 2^{2^n}$ for every natural $n$,...
- 2,878
3
votes
1
answer
226
views
Composite integers $n$ such that ALL factors of $2^n-1$ $\equiv$ $1$ $\pmod n$?
Are there any composite integers $n$ such that ALL factors of $2^n-1$ $\equiv$ $1$$\pmod n$. (In other words, every prime factor dividing $2^n-1$ has the form $2kn+1$) It seems unlikely that there ...
- 3,009
4
votes
1
answer
99
views
Show that there exist a prime divisor of $\sigma{((2^k)!)}$ which is greater than $2^k$
Let $k$ be a positive integer. Show that there exists a prime divisor of $\sigma{((2^k)!)}$ which is greater than $2^k$, where $\sigma{(n)}$ is the sum-of-divisors function.
user246688
0
votes
0
answers
77
views
Prove that if $x,y$ are coprime numbers then every odd factor of $x^{2^n}+y^{2^n}$ where $n$ is a positive integer , is of the form $2^{n+1}m+1$. [duplicate]
Prove that if $x$ and $y$ have no common factor then every odd factor of $x^{2^n}+y^{2^n}$ where $n$ is a positive integer , is of the form $2^{n+1}m+1$.
I have tried using order. Let $k$ be a odd ...
- 341
0
votes
0
answers
32
views
Quadratic residues and their using in proving the Édouard Lucas theorem related to Fermat numbers. [duplicate]
I was doing a study on Fermat Numbers when I came across this theorem by Édouard Lucas (unproven in my reference material):
Every prime divisor of $F_n = 2^{2^{n}} + 1$ is of the form $k \cdot{2^{n+2}...
- 148