All prime divisors of $\frac{x^m+1}{x+1}$ are of the form $2km+1$. Let $m$ be an odd prime and $x$ be the product of all primes of the form $2km+1$. Then all prime divisors of $\frac{x^m+1}{x+1}$ are of the form $2km+1$.
What I know is that $\frac{x^m+1}{x+1}$ is an integer.
Here is the link to the answer which prompted this question. 
Can anyone help me how to prove this.
Any help would be appreciated. Thanks in advance.
 A: We have $2 \nmid \frac{x^m+1}{x+1}$. Let odd prime $p$ divide $\frac{x^m+1}{x+1}$ :
Case $1$ : $m \mid (p-1)$
We clearly have $p=mq+1$ for some $q \in \mathbb{N}$. As $p$ is an odd prime, $mq+1$ is odd, and thus, $mq$ is even. Moreover, $m$ is an odd prime, thus, $q=2k$ for some $k \in \mathbb{N}$. Substituting:
$$p=2km+1$$
which proves that our prime divisor is of the required form.

Case $2$ : $m \nmid (p-1)$
We have:
$$p \mid (x^m+1) \implies p \mid(x^{2m}-1) \implies p \mid(x^{\gcd(2m,p-1)}-1)$$
by Fermat's Little Theorem.
Since $m$ is an odd prime not dividing $p-1$, it follows:
$$\gcd(2m,p-1)=\gcd(2,p-1)=2$$
This shows us that $p \mid (x^2-1)$.
We thus either have $p \mid (x-1)$ or $p \mid (x+1)$.
Subcase $1$ : $p \mid (x-1)$
We have:
$$p \mid (x-1) \implies p \mid (x^m-1)$$
Since $p \mid (x^m+1)$, it follows that $(x^m+1)-(x^m-1)=2$ is also divisible by $p$ which is a contradiction as $p$ is an odd prime.
Subcase $2$ : $p \mid (x+1)$
This is the same as $x \equiv -1 \pmod{p}$. But then:
$$\frac{x^m+1}{x+1} \equiv x^{m-1}-x^{m-2}+\cdots+1 \equiv 1-(-1)+1-(-1)+\cdots+1 \equiv m \pmod{p}$$
As $p \mid \frac{x^m+1}{x+1}$, it follows that $p \mid m$. Since $p$ and $m$ are both odd primes, we must thus have $p=m$. 
However:
$$p \mid (x^m+1) \implies m \mid (x^m+1)$$
Note that as all the prime factors of $x$ are $1 \pmod{m}$, we have $x \equiv 1 \pmod{m}$. Then:
$$0 \equiv x^m+1 \equiv 1+1 \equiv 2 \pmod{m} \implies m \mid 2$$
and this is once again a contradiction since $m$ is an odd prime.

Thus, we have proved that all prime divisors of $\frac{x^m+1}{x+1}$ are of the form $2km+1$.
A: As  $x$ is product of all primes of the form $2km+1$ it's also of this form. So, $x^m+1$ isn of this form $2Km+2$. Means $2m \nmid x^m+1$ as $m>2$.
If some prime $p >2, p \nmid {x+1}$ divides $\frac{x^m+1}{x+1}$, it also divides $x^m+1$.
So, $x^m \equiv -1 (\text{mod} p)$
or, $x^{2m} \equiv 1 (\text{mod} p)$.
Also, as $p|x^m+1, \text{gcd}(p, x)=1$
This implies that $x^{p-1} \equiv 1 (\text{mod} p)$. ( From Fermat's little theorem)
$\big [$ There doesn't exist any $2<r<2m$ such that $x^r \equiv 1 (\text{mod} p)$. Otherwise $r|2m \rightarrow r|m$. But $m$ is a prime.
Also $p \nmid x^2-1$ as $p \nmid x+1$. Because if some prime $q|x+1$, $q \nmid \frac{x^m+1}{x+1}$ ....(0) $\big ]$
$\Big ($ Proof to (0): If $q|(x+1)$ then , $x \equiv -1(\text{mod} q)$ (surely $q$ can't be equal to $m$). 
Now, $A=\frac{x^m+1}{x+1}=x^{m-1}-x^{m-2}+.....+(-1)^{m-2}1$ As, $m$ is odd prime, $A \equiv m (\text{mod} q)$. If  $A$ is divisible by $q$, then $q|m$. But that's impossible as $m$ is  an odd prime. $\Big)$

Hence, $2m|(p-1) \rightarrow p=2km+1$ for some $k \in \mathbb N$ 

Also the  case that $p-1|2m$ isn't possible being $m>1$ a prime and $p \geq 3$.
So all the prime factors dividing $\frac{x^m+1}{x+1}$ is of the form $2km+1$.
