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Let $p,n$ be two odd prime numbers. I want to show that the number of irreducible factors of $x^{p^n + 1} - 1$ over $\Bbb F_p$ is $$N = \frac{p^n-p}{2n} + \frac{p-1}{2} + 2$$

I know that this is equal to $\sum_{d\mid (p^n+1)}\frac{\phi(d)}{\operatorname{ord}_d(p)}$ (see How many irreducible factors does $x^n-1$ have over finite field?). I checked the equality with SAGE for some pairs $(p,n)$. But how would I prove the above equality?

Thank you!

Alphonse
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1 Answers1

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The number of irreducible factors of $X^m-1$ for $p\nmid m$, is the number of cycles of the Frobenius map $F:\alpha\to\alpha^p$ in its action on $\mu_m$, the set of $m$-th roots of unity in an algebraic closure of $\Bbb F_p$. This is the same as the number of orbits of $\phi:a\mapsto pa$ on $\Bbb Z/m\Bbb Z$.

In our case $m=p^n+1$. Then $\phi^n(a)=-a$ and so $\phi^{2n}(a)=a$. Every orbit of $\phi$ on $\Bbb Z/m\Bbb Z$ has length dividing $2n$. For $k\mid 2n$, the number of solutions to $\phi^k(a)=a$ is the number of solutions to the congruence $(p^k-1)a\equiv0\pmod{p^n+1}$ and equals $a_k=\gcd(p^k-1,p^n+1)$.

Now $a_1=\gcd(p-1,p^n+1)=\gcd(p-1,p+1)=2$. Also $a_2=\gcd(p^2-1,p^n+1)=\gcd(p^2-1,p+1)=p+1$ as $p^n\equiv p\pmod{p^2}$. Finally $a_n=\gcd(p^n-1,p^n+1)=\gcd(p^n-1,2)=2$.

The orbits of length $1$ comprise two elements, those of length $2$ comprise $(p+1)-2 = p-1$ elements and there are no orbits of length $n$. Therefore the orbits of length $2n$ comprise $$p^n+1-2-(p-1)=p^n-p$$ elements and so there are $2$ orbits of length $1$, $(p-1)/2$ of length $2$ and $$\frac{p^n-p}{2n}$$ of length $2n$. The number of irreducible factors is the total number of orbits, namely $$\frac{p^n-p}{2n}+\frac{p-1}2+2.$$

Angina Seng
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