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.
Best Answer
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 $2km+1$.